PDF 8: Quantities In Chemical Reactions

8: QUANTITIES IN CHEMICAL REACTIONS  CHAPTER OVERVIEW 8: Quantities in Chemical Reactions So far, we have talked about chemical reactions in terms of individual atoms and molecules. Although this works, most of the reactions occurring around us involve much larger amounts of chemicals. Even a tiny...

8: QUANTITIES IN CHEMICAL REACTIONS  CHAPTER OVERVIEW 8: Quantities in Chemical Reactions So far, we have talked about chemical reactions in terms of individual atoms and molecules. Although this works, most of the reactions occurring around us involve much larger amounts of chemicals. Even a tiny sample of a substance will contain millions, billions, or a hundred billion billions of atoms and molecules. How do we compare amounts of substances to each other, in chemical terms, when it is so difficult to count to a hundred billion billion? Actually, there are ways to do this, which we will explore in this chapter. In doing so, we will increase our understanding of stoichiometry, which is the study of the numerical relationships between the reactants and the products in a balanced chemical reaction. 8.1: Climate Change - Too Much Carbon Dioxide 8.2: Making Pancakes- Relationships Between Ingredients 8.3: Making Molecules- Mole-to-Mole Conversions 8.4: Making Molecules- Mass-to-Mass Conversions 8.5: Stoichiometry 8.6: Limiting Reactant and Theoretical Yield 8.7: Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Masses of Reactants 8.8: Enthalpy Change is a Measure of the Heat Evolved or Absorbed 8: Quantities in Chemical Reactions is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 1 8.1: CLIMATE CHANGE - TOO MUCH CARBON DIOXIDE Carbon dioxide (CO2) is an important heat-trapping (greenhouse) gas, which is released through human activities such as deforestation and burning fossil fuels, as well as natural processes such as respiration and volcanic eruptions. Figure 8.1.1 shows CO2 levels during the last three glacial cycles, as reconstructed from ice cores. Figure 8.1.1: This graph, based on the comparison of atmospheric samples contained in ice cores and more recent direct measurements, provides evidence that atmospheric CO2 has increased since the Industrial Revolution. (Luthi, D., et al. 2008; Etheridge, D.M., et al. 2010; Line graph of carbon dioxide levels Vostok ice core data/J.R.R. Petit et al.; NOAA Mauna Loa CO2 record. NASA.) from 800 thousand before up until present day; current carbon dioxide level is higher than all historical levels. Carbon dioxide (CO ) is the primary greenhouse gas emitted through human activities. In 2015, CO accounted for about 82.2% of all U.S. 2 2 greenhouse gas emissions from human activities. Carbon dioxide is naturally present in the atmosphere as part of the Earth's carbon cycle (the natural circulation of carbon among the atmosphere, oceans, soil, plants, and animals). Human activities are altering the carbon cycle, both by adding more CO to the atmosphere and by influencing the ability of natural sinks, like forests, to remove CO from the 2 2 atmosphere. While CO emissions come from a variety of natural sources, human-related emissions are responsible for the increase that has 2 occurred in the atmosphere since the industrial revolution. The main human activity that emits CO is the combustion of fossil fuels (coal, natural gas, and oil) for energy and transportation, although 2 certain industrial processes and land-use changes also emit CO. As an example of how CO can be generated, consider the combustion of 2 2 octane, a component of gasoline: 2C H (l) + 21 O (g) → 16 CO (g) + 18 H O(g) (8.1.1) 8 18 2 2 2 The balanced reaction in Equation 8.1.1 demonstrates that for every two molecules of octane that are burned, 16 molecules of CO2 are generated. CONTRIBUTIONS & ATTRIBUTIONS Earth Science Communications Team at NASA's Jet Propulsion Laboratory, California Institute of Technology EPA 8.1: Climate Change - Too Much Carbon Dioxide is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Marisa Alviar- Agnew & Henry Agnew. 8.1.1 https://chem.libretexts.org/@go/page/47497 8.2: Making Pancakes- Relationships Between Ingredients 8.2: Making Pancakes- Relationships Between Ingredients is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. 8.2.1 https://chem.libretexts.org/@go/page/396613 8.3: MAKING MOLECULES- MOLE-TO-MOLE CONVERSIONS  /;LEARNING OBJECTIVES Use a balanced chemical equation to determine molar relationships between substances. Previously, you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). As follows, we will extend the meaning of the coefficients in a chemical equation. Consider the simple chemical equation: 2H2 + O2 → 2H2 O The convention for writing balanced chemical equations is to use the lowest whole-number ratio for the coefficients. However, the equation is balanced as long as the coefficients are in a 2:1:2 ratio. For example, this equation is also balanced if we write it as 4H2 + 2O2 → 4H2 O The ratio of the coefficients is 4:2:4, which reduces to 2:1:2. The equation is also balanced if we were to write it as 22H2 + 11O2 → 22H2 O because 22:11:22 also reduces to 2:1:2. Suppose we want to use larger numbers. Consider the following coefficients: 23 23 23 12.044 × 10 H + 6.022 × 10 O → 12.044 × 10 H O 2 2 2 These coefficients also have the ratio 2:1:2 (check it and see), so this equation is balanced. But 6.022 × 1023 is 1 mol, while 12.044 × 1023 is 2 mol (and the number is written that way to make this more obvious), so we can simplify this version of the equation by writing it as 2 mol H + 1 mol O → 2 mol H O 2 2 2 We can leave out the word mol and not write the 1 coefficient (as is our habit), so the final form of the equation, still balanced, is 2H +O → 2H O 2 2 2 Now we interpret the coefficients as referring to molar amounts, not individual molecules. The lesson? Balanced chemical equations are balanced not only at the molecular level, but also in terms of molar amounts of reactants and products. Thus, we can read this reaction as “two moles of hydrogen react with one mole of oxygen to produce two moles of water.” By the same token, the ratios we constructed to describe a molecular reaction can also be constructed in terms of moles rather than molecules. For the reaction in which hydrogen and oxygen combine to make water, for example, we can construct the following ratios: 2 mol H2 1 mol O2 or 1 mol O2 2 mol H2 2 mol H2 O 1 mol O2 or 1 mol O2 2 mol H2 O 2 mol H2 2 mol H2 O or 2 mol H2 O 2 mol H2 We can use these ratios to determine what amount of a substance, in moles, will react with or produce a given number of moles of a different substance. The study of the numerical relationships between the reactants and the products in balanced chemical reactions is called stoichiometry.  EXAMPLE 8.3.1 How many moles of oxygen react with hydrogen to produce 27.6 mol of H 2 O ? Solution 8.3.1 https://chem.libretexts.org/@go/page/47501  Solutions to Example 8.3.1 Steps for Problem Solving How many moles of oxygen react with hydrogen to produce 27.6 mol of H 2 O ? Unbalanced: H2 + O2 → H2O Find a balanced equation that describes the reaction. Balanced: 2H2 + O2 → 2H2O Identify the "given" information and what the problem is Given: moles H2O Find: moles oxygen asking you to "find." List other known quantities. 1 mol O2 = 2 mol H2O Prepare a concept map and use the proper conversion factor. 1 mol O2 27.6 mol H2 O × = 13.8 mol O2 Cancel units and calculate. 2 mol H2 O To produce 27.6 mol of H2O, 13.8 mol of O2 react. Since each mole of oxygen produces twice as many moles of water, it makes sense that the produced Think about your result. amount is greater than the reactant amount  EXAMPLE 8.3.2 How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen? Solution Solutions to Example 8.3.2 How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an Steps for Problem Solving excess of nitrogen? Unbalanced: N2 + H2 → NH3 Find a balanced equation that describes the reaction. Balanced: N2 + 3H2 → 2NH3 Identify the "given" information and what the problem is asking Given: H = 4.20 mol 2 you to "find." Find: mol of NH 3 List other known quantities. 3 mol H2 = 2 mol NH3 Prepare a concept map and use the proper conversion factor. 2 mol N H3 4.20 mol H2 × = 2.80 mol N H3 Cancel units and calculate. 3 mol H2 The reaction of 4.20 mol of hydrogen with excess nitrogen produces 2.80 mol of ammonia. Think about your result. The result corresponds to the 3:2 ratio of hydrogen to ammonia from the balanced equation.  EXERCISE 8.3.3 a. Given the following balanced chemical equation: C H +8O → 5 CO +6H O 5 12 2 2 2 , How many moles of H O can be formed if 0.0652 mol of C H were to react? 2 5 12 b. Balance the following unbalanced equation and determine how many moles of H 2 O are produced when 1.65 mol of NH3 react: NH +O → N +H O 3 2 2 2 Answer a 0.391 mol H2O 8.3.2 https://chem.libretexts.org/@go/page/47501  Answer b 4NH3 + 3O2 → 2N2 + 6H2O; 2.48 mol H2O SUMMARY The balanced chemical reaction can be used to determine molar relationships between substances. 8.3: Making Molecules- Mole-to-Mole Conversions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 8.3.3 https://chem.libretexts.org/@go/page/47501 8.4: MAKING MOLECULES- MASS-TO-MASS CONVERSIONS  LEARNING OBJECTIVES Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction. MOLE TO MASS CONVERSIONS We have established that a balanced chemical equation is balanced in terms of moles, as well as atoms or molecules. We have used balanced equations to set up ratios, in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions—such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that relation to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence: Collectively, these conversions are called mole-mass calculations. Flowchart of mole mass calculations: To convert from moles of substance A to moles of substance B, use the mole ratio conversion factor, and to convert from moles to grams of substance B, use molar mass conversion factor As an example, consider the balanced chemical equation Fe O + 3 SO → Fe (SO ) (8.4.1) 2 3 3 2 4 3 If we have 3.59 mol of Fe O , how many grams of SO can react with it? Using the mole-mass calculation sequence, we can determine the 2 3 3 required mass of SO in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to 3 calculate the number of moles of SO needed. Then, using the molar mass of SO as a conversion factor, we determine the mass that this 3 3 number of moles of SO has. 3 As usual, we start with the quantity we were given: 3 mol SO3 3.59 mol Fe2 O3 ×( ) = 10.77 mol SO3 (8.4.2) 1 mol Fe2 O3 The mol Fe O units cancel, leaving mol SO unit. Now, we take this answer and convert it to grams of SO , using the molar mass of SO 2 3 3 3 3 as the conversion factor: 80.06 g SO3 10.77 mol SO3 ×( ) = 862 g SO3 (8.4.3) 1 mol SO3 Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO will react with 3.59 mol of 3 Fe O. Many problems of this type can be answered in this manner. 2 3 The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows: 8.4.1 https://chem.libretexts.org/@go/page/47503  3 mol S O3 80.06 g S O3 3.59 mol F e2 O3 × ( )× ( ) = 862 g S O3 1 mol F e2 O3 1 mol S O3   converts to moles of SO 3 converts to grams of SO 3 We get exactly the same answer when combining all math steps together.  EXAMPLE 8.4.1: GENERATION OF ALUMINUM OXIDE How many moles of HCl will be produced when 249 g of AlCl are reacted according to this chemical equation? 3 2 AlCl + 3 H O(ℓ) → Al O + 6 HCl(g) 3 2 2 3 Solution Solutions to Example 8.5.1 Steps for Problem Solving Example 8.4.1 Identify the "given" information and what the Given: 249 g AlCl3 problem is asking you to "find." Find: moles HCl 1 mol AlCl3 = 133.33 g AlCl3 List other known quantities. 6 mol of HCl to 2 mol AlCl3 Prepare a concept map and use the proper conversion factor. Flowchart of needed conversion factors: 1 mole AlCl3 to 133.33 grams AlCl3, and 6 moles HCl to 2 moles AlCl3 1 mol AlCl3 6 mol HCl Cancel units and calculate. 249 g AlCl3 × × = 5.60 mol HCl 133.33 g AlCl3 2 mol AlCl3 Since 249 g of AlCl3 is less than 266.66 g, the mass for 2 moles of AlCl3 and the relationship is 6 mol of HCl to 2 Think about your result. mol AlCl3 , the answer should be less than 6 moles of HCl.  EXERCISE 8.4.1: GENERATION OF ALUMINUM OXIDE How many moles of Al 2 O3 will be produced when 23.9 g of H 2 O are reacted according to this chemical equation? 2 AlCl + 3 H O(ℓ) → Al O + 6 HCl(g) 3 2 2 3 Answer 0.442 mol Al2O3 MASS TO MASS CONVERSIONS It is a small step from mole-mass calculations to mass-mass calculations. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then—and only then—we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows: 8.4.2 https://chem.libretexts.org/@go/page/47503 This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques. Flowchart of mole mass calculations: To convert from grams to moles of substance A, use molar mass conversion factor; To convert from moles of substance A to moles of substance B, use the mole ratio conversion factor, and to convert from moles to grams of substance B, use molar mass conversion factor  EXAMPLE 8.4.2: DECOMPOSITION OF AMMONIUM NITRATE Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation. NH NO (s) → N O (g) + 2H O (l) 4 3 2 2 In a certain experiment, 45.7 g of ammonium nitrate is decomposed. Find the mass of each of the products formed. 8.4.3 https://chem.libretexts.org/@go/page/47503  Solutions to Example 8.5.2 Steps for Problem Solving Example 8.4.2 Given: 45.7 g NH 4 NO 3 Identify the "given" information and what the Find: problem is asking you to "find." Mass N O =? g 2 Mass H O =? g 2 1 mol NH NO = 80.06 g 4 3 1 mol N O = 44.02 g List other known quantities. 2 1 mol H O = 18.02 g 2 1 mol NH4NO3 to 1 mol N2O to 2 mol H2O Flowchart of conversion factors: 1 mole NH4NO3 to 80.06 grams NH4NO3, 1 mole N2O to 1 mole NH4NO3, 44.02 Prepare two concept maps and use the proper grams N2O to 1 mole N2O conversion factor. Flowchart of conversion factors: 1 mole NH4NO3 to 80.06 grams NH4NO3, 2 moles H2O to 1 mole NH4NO3, 18.02 grams H2O to 1 mole H2O 1 mol NH NO 1 mol N O 44.02 g N O 4 3 2 2 45.7 g NH NO × × × = 25.1 g N O 4 3 2 80.06 g NH NO 1 mol NH NO 1 mol N O Cancel units and calculate. 4 3 4 3 2 1 mol NH NO 2 mol H O 18.02 g H O 4 3 2 2 45.7 g NH NO × × × = 20.6 g H O 4 3 2 80.06 g NH NO 1 mol NH NO 1 mol H O 4 3 4 3 2 The total mass of the two products is equal to the mass of ammonium nitrate which decomposed, demonstrating the Think about your result. law of conservation of mass. Each answer has three significant figures.  EXERCISE 8.4.2: CARBON TETRACHLORIDE Methane can react with elemental chlorine to make carbon tetrachloride (CCl ). The balanced chemical equation is as follows: 4 CH (g) + 4 Cl (g) → CCl (l) + 4 HCl(l) 4 2 2 How many grams of HCl are produced by the reaction of 100.0 g of CH ? 4 Answer 908.7g HCl SUMMARY Calculations involving conversions between moles of a substance and the mass of that substance can be done using conversion factors. A balanced chemical reaction can be used to determine molar and mass relationships between substances. 8.4: Making Molecules- Mass-to-Mass Conversions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 8.4.4 https://chem.libretexts.org/@go/page/47503 8.5: Stoichiometry  Learning Objectives Explain the meaning of the term "stoichiometry". Determine the relative amounts of each substance in chemical equations. You have learned that chemical equations provide us with information about the types of particles that react to form products. Chemical equations also provide us with the relative number of particles and moles that react to form products. In this section you will explore the quantitative relationships that exist between the quantities of reactants and products in a balanced equation. This is known as stoichiometry. Stoichiometry, by definition, is the calculation of the quantities of reactants or products in a chemical reaction using the relationships found in the balanced chemical equation. The word stoichiometry is actually Greek from two words: στ οικηιον, which means "element", and \(\mu \epsilon \tau \rho \omicron \nu), which means "measure". Interpreting Chemical Equations The mole, as you remember, is a quantitative measure that is equivalent to Avogadro's number of particles. So how does this relate to the chemical equation? Look at the chemical equation below. 2 CuSO + 4KI → 2CuI + 2 K SO +I 4 2 4 2 The coefficients used, as we have learned, tell us the relative amounts of each substance in the equation. So for every 2 units of copper (II) sulfate (CuSO ) we have, we need to have 4 units of potassium iodide (KI ). For every two dozen copper (II) sulfates, 4 we need 4 dozen potassium iodides. Because the unit "mole" is also a counting unit, we can interpret this equation in terms of moles, as well: For every two moles of copper (II) sulfate, we need 4 moles potassium iodide. The production of ammonia (NH 3 ) from nitrogen and hydrogen gases is an important industrial reaction called the Haber process, after German chemist Fritz Haber. N (g) + 3 H (g) → 2 NH (g) 2 2 3 The balanced equation can be analyzed in several ways, as shown in the figure below. Figure 8.5.1 : This representation of the production of ammonia from nitrogen and hydrogen show several ways to interpret the quantitative information of a chemical reaction. We see that 1 molecule of nitrogen reacts with 3 molecules of hydrogen to form 2 molecules of ammonia. This is the smallest possible relative amount of the reactants and products. To consider larger relative amounts, each coefficient can be multiplied by the same number. For example, 10 molecules of nitrogen would react with 30 molecules of hydrogen to produce 20 molecules of ammonia. The most useful quantity for counting particles is the mole. So if each coefficient is multiplied by a mole, the balanced chemical equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This is the conventional way to interpret any balanced chemical equation. Finally, if each mole quantity is converted to grams by using the molar mass, we can see that the law of conservation of mass is followed. 1 mol of nitrogen has a mass of 28.02 g, while 3 mol of hydrogen has a mass of 6.06 g, and 2 mol of ammonia has a mass of 34.08 g. 8.5.1 https://chem.libretexts.org/@go/page/47499  28.02 g N + 6.06 g H → 34.08 g NH 2 2 3 Mass and the number of atoms must be conserved in any chemical reaction. The number of molecules is not necessarily conserved.  Example 8.5.1 The equation for the combustion of ethane (C 2 H 6 ) is 2C H + 7O → 4 CO + 6H O 2 6 2 2 2 a. Indicate the number of formula units or molecules in the balanced equation. b. Indicate the number of moles present in the balanced equation. Solution a. Two molecules of C H plus seven molecules of O yields four molecules of CO plus six molecules of H 2 6 2 2 2 O. b. Two moles of C H plus seven moles of O yields four moles of CO plus six moles of H O. 2 6 2 2 2  Exercise 8.5.1 For the following equation below, indicate the number of formula units or molecules, and the number of moles present in the balanced equation. KBrO + 6KI + 6HBr → 7KBr + 3 H O 3 2 Answer One molecules of KBrO plus six molecules of KI plus six molecules of HBr yields seven molecules of KBr plus three 3 molecules of I and three molecules of H O. One mole of KBrO plus six moles of KI plus six moles of HBr yields 2 2 3 seven moles of KBr plus three moles of I plus three moles of H O. 2 2 Summary Stoichiometry is the calculation of the quantities of reactants or products in a chemical reaction using the relationships found in a balanced chemical equation. The coefficients in a balanced chemical equation represent the reacting ratios of the substances in the reaction. The coefficients of a balanced equation can be used to determine the ratio of moles of all substances in the reaction. Vocabulary Stoichiometry - The calculation of quantitative relationships of the reactants and products in a balanced chemical equation. Formula unit - The empirical formula of an ionic compound. Mole ratio - The ratio of the moles of one reactant or product to the moles of another reactant or product according to the coefficients in the balanced chemical equation. 8.5: Stoichiometry is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 8.5.2 https://chem.libretexts.org/@go/page/47499 8.6: LIMITING REACTANT AND THEORETICAL YIELD  LEARNING OBJECTIVES Identify the limiting reactant (limiting reagent) in a given chemical reaction. Calculate how much product will be produced from the limiting reactant. Calculate how much reactant(s) remains when the reaction is complete. In all examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants were left over at the end of the reaction. This is often desirable—as in the case of a space shuttle—where excess oxygen or hydrogen is not only extra freight to be hauled into orbit, but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely, but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactant. The reactant that remains after a reaction has gone to completion is in excess. Consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is: 1 box mix + 2 eggs → 1 batch brownies (8.6.1) If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant. Even if you had a refrigerator full of eggs, you could make only two batches of brownies. Figure 8.6.1: The Concept of a Limiting Reactant in the Preparation of Brownies. For a chemist, the balanced chemical equation is the recipe that must be followed. 2 boxes of brownie mix and 12 eggs results in 2 batches of brownies and 8 eggs; in this case the 8 eggs are reactant present in excess  PHET SIMULATION: REACTANTS, PRODUCTS AND LEFTOVERS View this interactive simulation illustrating the concepts of limiting and excess reactants. Reactants, Products and Leftovers Molecules Game Sandwiches Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride: H + Cl (g) → 2 HCl(g) 2 2 The balanced equation shows that hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H2 8.6.1 https://chem.libretexts.org/@go/page/47505 and 2 moles of Cl2. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen non-reacted. An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example, in the previous paragraph, complete reaction of the hydrogen would yield: 2 mol HCl mol HCl produced = 3 mol H2 × = 6 mol HCl 1 mol H2 Complete reaction of the provided chlorine would produce: 2 mol HCl mol HCl produced = 2 mol Cl2 × = 4 mol HCl 1 mol Cl2 The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be non-reacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure 8.6.2 ). Figure 8.6.2: When H2 and Cl2 are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant. The figure shows a space-filling molecular models reacting. There is a reaction arrow pointing to the right in the middle. To the left of the reaction arrow there are three molecules each consisting of two green spheres bonded together. There are also five molecules each consisting of two smaller, white spheres bonded together. Above these molecules is the label, “Before reaction,” and below these molecules is the label, “6 H subscript 2 and 4 C l subscript 2.” To the right of the reaction arrow, there are eight molecules each consisting of one green sphere bonded to a smaller white sphere. There are also two molecules each consisting of two white spheres bonded together. Above these molecules is the label, “After reaction,” and below these molecules is the label, “8 H C l and 2 H subscript 2.” A similar situation exists for many chemical reactions: you usually run out of one reactant before all of the other reactant has reacted. The reactant you run out of is called the limiting reactant; the other reactant or reactants are considered to be in excess. A crucial skill in evaluating the conditions of a chemical process is to determine which reactant is the limiting reactant and which is in excess.  HOW TO IDENTIFY THE LIMITING REACTANT (LIMITING REAGENT) There are two ways to determine the limiting reactant. One method is to find and compare the mole ratio of the reactants used in the reaction (Approach 1). Another way is to calculate the grams of products produced from the given quantities of reactants; the reactant that produces the smallest amount of product is the limiting reactant (Approach 2). This section will focus more on the second method. Approach 1 (The "Reactant Mole Ratio Method"): Find the limiting reactant by looking at the number of moles of each reactant. 1. Determine the balanced chemical equation for the chemical reaction. 2. Convert all given information into moles (most likely, through the use of molar mass as a conversion factor). 3. Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio. 4. Use the amount of limiting reactant to calculate the amount of product produced. 5. If necessary, calculate how much is left in excess of the non-limiting (excess) reactant. Approach 2 (The "The Product Method"): Find the limiting reactant by calculating and comparing the amount of product that each reactant will produce. 1. Balance the chemical equation for the chemical reaction. 2. Convert the given information into moles. 3. Use stoichiometry for each individual reactant to find the mass of product produced. 4. The reactant that produces a lesser amount of product is the limiting reactant. 5. The reactant that produces a larger amount of product is the excess reactant. 6. To find the amount of remaining excess reactant, subtract the mass of excess reactant consumed from the total mass of excess reactant given. The key to recognizing which reactant is the limiting reactant is based on a mole-mass or mass-mass calculation: whichever reactant gives the lesser amount of product is the limiting reactant. What we need to do is determine an amount of one product (either moles or mass) assuming all of each reactant reacts. Whichever reactant gives the least amount of that particular product is the limiting reactant. It does not matter which product we use, as long as we use the same one each time. It does not matter whether we determine the number of moles or grams of that product; however, we will see shortly that knowing the final mass of product can be useful.  EXAMPLE 8.6.1: IDENTIFYING THE LIMITING REACTANT As an example, consider the balanced equation 4 C H Br + 11 O → 8 CO + 6 H O + 6 Br 2 3 3 2 2 2 2 What is the limiting reactant if 76.4 grams of C 2 H Br 3 3 reacted with 49.1 grams of O ? 2 Solution Using Approach 1: Step 1: Balance the chemical equation. The equation is already balanced with the relationship 4 mol C 2 H3 Br3 to 11 mol O to 6 mol H 2 2 O to 6 mol Br Step 2: Convert all given information into moles. 1 mol C2 H3 Br3 76.4 g C2 H3 Br3 × = 0.286 mol C2 H3 Br3 266.72 g C2 H3 B3 1 mol O2 49.1 g O2 × = 1.53 mol O2 32.00 g O2 Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio. Assuming that all of the oxygen is used up, 4 molC2 H3 Br3 1.53 molO2 × = 0.556 mol C2H3Br3 are required. 11 molO2 Because 0.556 moles of C2H3Br3 required > 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reactant. Using Approach 2: 8.6.2 https://chem.libretexts.org/@go/page/47505 Step 1: Balance the chemical equation. The equation is already balanced with the relationship 4 mol C2H3Br3 to 11 mol O2 to 6 mol H2O to 6 mol Br2 Step 2 and Step 3: Convert mass to moles and stoichiometry. 1 mol C2 H3 Br3 8 mol CO2 44.01 g CO2 76.4 g C2 H3 Br3 × × × = 25.2 g CO2 266.72 g C2 H3 Br3 4 mol C2 H3 Br3 1 mol CO2 1 mol O2 8 mol CO2 44.01 g CO2 49.1 g O2 × × × = 49.1 g CO2 32.00 g O2 11 mol O2 1 mol CO2 Step 4: The reactant that produces a smaller amount of product is the limiting reactant. Therefore, by either method, C 2 H Br 3 3 is the limiting reactant.  EXAMPLE 8.6.2: IDENTIFYING THE LIMITING REACTANT AND THE MASS OF EXCESS REACTANT For example, in the reaction of magnesium metal and oxygen, calculate the mass of magnesium oxide that can be produced if 2.40 g M g reacts with 10.0 g O. Also determine the amount of excess reactant. 2 MgO is the only product in the reaction. Solution Following Approach 1: Step 1: Balance the chemical equation. 2 Mg (s) + O2 (g) → 2 MgO (s) The balanced equation provides the relationship of 2 mol Mg to 1 mol O2 to 2 mol MgO Step 2 and Step 3: Convert mass to moles and stoichiometry. 1 mol Mg 2 mol MgO 40.31 g MgO 2.40 g Mg × × × = 3.98 g MgO 24.31 g Mg 2 mol Mg 1 mol MgO 1 mol O2 2 mol MgO 40.31 g MgO 10.0 g O2 × × × = 25.2 g MgO 32.00 g O2 1 mol O2 1 mol MgO Step 4: The reactant that produces a smaller amount of product is the limiting reactant. Mg produces less MgO than does O2 (3.98 g MgO vs. 25.2 g MgO), therefore Mg is the limiting reactant in this reaction. Step 5: The reactant that produces a larger amount of product is the excess reactant. O2 produces more amount of MgO than Mg (25.2g MgO vs. 3.98 MgO), therefore O2 is the excess reactant in this reaction. Step 6: Find the amount of remaining excess reactant by subtracting the mass of the excess reactant consumed from the total mass of excess reactant given. Mass of excess reactant calculated using the limiting reactant: 1 mol Mg 1 mol O2 32.00 g O2 2.40 g Mg × × × = 1.58 g O2 24.31 g Mg 2 mol Mg 1 mol O2 OR Mass of excess reactant calculated using the mass of the product: 1 mol MgO 1 mol O2 32.0 g O2 3.98 g MgO × × × = 1.58 g O2 40.31 g MgO 2 mol MgO 1 mol O2 Mass of total excess reactant given – mass of excess reactant consumed in the reaction: 10.0g O2 - (available) 1.58g O2 (used) = 8.42g O2 (excess) Therefore, O2 is in excess.  EXAMPLE 8.6.3: LIMITING REACTANT What is the limiting reactant if 78.0 grams of Na2O2 were reacted with 29.4 grams of H2O? The unbalanced chemical equation is Na 2 O2 (s) + H2 O(l) → NaOH(aq) + H2 O2 (l) Solution 8.6.3 https://chem.libretexts.org/@go/page/47505  Solutions to Example 8.4.3 Steps for Problem Solving- The Product Method Example 8.6.1 Given: 78.0 grams of Na2O2 Identify the "given" information and what the problem is asking you to "find." 29.4 g H2O Find: limiting reactant 1 mol Na2O2= 77.96 g/mol List other known quantities. 1 mol H2O = 18.02 g/mol Since the amount of product in grams is not required, only the molar mass of the reactants is needed. Na2O2 (s) + 2H2O (l) → 2NaOH (aq) + H2O2 (l) Balance the equation. The balanced equation provides the relationship of 1 mol Na2O2 to 2 mol H2O 2mol NaOH to 1 mol H2O2 Prepare a concept map and use the proper conversion factor. Because the question only asks for the limiting reactant, we can perform two mass-mole calculations and determine which amount is less. 1 mol Na2 O2 2 mol NaOH 40 g NaOH 78.0 g N a2 O2 × × × = 2.00 mol NaOH 77.96 g Na2 O2 1 mol Na2 O2 1 mol NaOH Cancel units and calculate. 1 mol H2 O 2 mol NaOH 40 g NaOH 29.4 g H2 O × × × = 1.63 mol NaOH 18.02 g H2 O 2 mol Na2 O2 1 mol NaOH Therefore, H2O is the limiting reactant. Think about your result.  EXAMPLE 8.6.4: LIMITING REACTANT AND MASS OF EXCESS REACTANT A 5.00 g quantity of Rb is combined with 3.44 g of MgCl according to this chemical reaction: 2 2Rb(s) + M gC l2 (s) → M g(s) + 2RbCl(s) What mass of Mg is formed, and what mass of remaining reactant is left over? Solution 8.6.4 https://chem.libretexts.org/@go/page/47505  Solutions to Example 8.4.10 Steps for Problem Solving- The Product Example 8.6.2 Method A 5.00 g quantity of Rb is combined with 3.44 g of MgCl2 according to this chemical reaction: Steps for Problem Solving 2Rb(s) + M gC l2 (s) → M g(s) + 2RbCl(s) What mass of Mg is formed, and what mass of remaining reactant is left over? Identify the "given" information and what the Given: 5.00g Rb, 2.44g MgCl2 problem is asking you to "find." Find: mass of Mg formed, mass of remaining reactant molar mass: Rb = 85.47 g/mol List other known quantities. molar mass: MgCl2 = 95.21 g/mol molar mass: Mg = 24.31 g/mol Find mass Mg formed based on mass of Rb Find mass of Mg formed based on mass of MgCl2 Prepare concept maps and use the proper conversion factor. Conversion factors: 1 mole MgCl2 to 95.21 grams MgCl2, 1 mole Mg to 1 mole MgCl2, 24.31 grams Mg to 1 mole Mg Use limiting reactant to determine amount of excess reactant consumed Conversion factors: 1 mole Rb to 85.47 grams Rb, 1 mole MgCl2 to 2 moles Rb, 95.21 grams MgCl2 to 1 mole MgCl2 Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less. 1 mol Rb 1 mol M g 24.31 g M g 5.00 g Rb × × × = 0.711 g M g 85.47 g Rb 2 mol Rb 1 mol M g 1 mol M gCl2 1 mol M g 24.31 g M g 3.44 g M gCl2 × × × = 0.878 g M g 95.21 g M gCl2 1 mol M gCl2 1 mol M g Cancel units and calculate. The 0.711 g of Mg is the lesser quantity, so the associated reactant—5.00 g of Rb—is the limiting reactant. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl2 reacted with the 5.00 g of Rb, and then subtract the amount reacted from the original amount. 1 mol Rb 1 mol M gCl2 95.21 g M gCl2 5.00 g Rb × × × = 2.78 g M gC l2 reacted 85.47 g Rb 2 mol Rb 1 mol M gCl2 Because we started with 3.44 g of MgCl2, we have 3.44 g MgCl2 − 2.78 g MgCl2 reacted = 0.66 g MgCl2 left Think about your result. It usually is not possible to determine the limiting reactant using just the initial masses, as the reagents have different molar masses and coefficients.  EXERCISE 8.6.1 Given the initial amounts listed, what is the limiting reactant, and what is the mass of the leftover reactant? 22.7 g + 17.9 g → M gS (s) + H2 O(l)   M gO(s) H2 S Answer H2S is the limiting reagent; 1.5 g of MgO are left over. 8.6: Limiting Reactant and Theoretical Yield is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Sarick Shah, Marisa Alviar-Agnew, Henry Agnew, & Henry Agnew. 8.6.5 https://chem.libretexts.org/@go/page/47505 8.6.6 https://chem.libretexts.org/@go/page/47505 8.6.7 https://chem.libretexts.org/@go/page/47505 8.7: LIMITING REACTANT, THEORETICAL YIELD, AND PERCENT YIELD FROM INITIAL MASSES OF REACTANTS  LEARNING OBJECTIVES Calculate percentage or actual yields from known amounts of reactants. The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Studying how much of a compound is produced in any given reaction is an important part of cost control. PERCENT YIELD Chemical reactions in the real world don't always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than predicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield. To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that can be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage. Actual Yield Percent Yield = × 100% Theoretical Yield Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense. Typically, percent yields are understandably less than 100% because of the reasons indicated earlier. However, percent yields greater than 100% are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction. Example 8.7.1 illustrates the steps for determining percent yield.  EXAMPLE 8.7.1: DECOMPOSITION OF POTASSIUM CHLORATE Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction below: 2KClO3 (s) → 2KCl (s) + 3O2 (g) In a certain experiment, 40.0 g KClO is heated until it completely decomposes. The experiment is performed and the oxygen gas is 3 collected and its mass is found to be 14.9 g. a. What is the theoretical yield of oxygen gas? b. What is the percent yield for the reaction? Solution a. Calculation of theoretical yield First, we will calculate the theoretical yield based on the stoichiometry. Step 1: Identify the "given" information and what the problem is asking you to "find". Given: Mass of KClO 3 = 40.0 g Mass of O2 collected = 14.9g Find: Theoretical yield, g O2 Step 2: List other known quantities and plan the problem. 1 mol KClO3 = 122.55 g/mol 1 mol O2 = 32.00 g/mol Step 3: Apply stoichiometry to convert from the mass of a reactant to the mass of a product: 8.7.1 https://chem.libretexts.org/@go/page/47507 Step 4: Solve. 1 mol KClO 3 mol O 3 2 32.00 g O 2 40.0 g KClO × × × = 15.7 g O 3 2 122.55 g KClO3 2 mol KClO3 1 mol O2 The theoretical yield of O is 15.7 g , 15.67 g unrounded. 2 Step 5: Think about your result. The mass of oxygen gas must be less than the 40.0 g of potassium chlorate that was decomposed. b. Calculation of percent yield Now we will use the actual yield and the theoretical yield to calculate the percent yield. Step 1: Identify the "given" information and what the problem is asking you to "find". Given: Theoretical yield =15.67 g, use the un-rounded number for the calculation. Actual yield = 14.9g Find: Percent yield, % Yield Step 2: List other known quantities and plan the problem. No other quantities needed. Step 3: Use the percent yield equation below. Actual Yield Percent Yield = × 100% Theoretical Yield Step 4: Solve. 14.9 g Percent Yield = × 100% = 94.9% 15.67 g – Step 5: Think about your result. Since the actual yield is slightly less than the theoretical yield, the percent yield is just under 100%.  EXAMPLE 8.7.2: OXIDATION OF ZINC Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation: CuSO (aq) + Zn(s) → Cu(s) + ZnSO (aq) 4 4 What is the percent yield? Solution 8.7.2 https://chem.libretexts.org/@go/page/47507  Solutions to Example 8.6.2 Steps for Problem Solving-The Product Example 8.7.1 Method Given: 1.274 g CuSO4 Identify the "given" information and what the Actual yield = 0.392 g Cu problem is asking you to "find." Find: Percent yield 1 mol CuSO4= 159.62 g/mol List other known quantities. 1 mol Cu = 63.55 g/mol Since the amount of product in grams is not required, only the molar mass of the reactants is needed. The chemical equation is already balanced. Balance the equation. The balanced equation provides the relationship of 1 mol CuSO4 to 1 mol Zn to 1 mol Cu to 1 mol ZnSO4. Prepare a concept map and use the proper conversion factor. The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield (g Cu) is found by performing mass-mass calculation based on the initial amount of CuSO4. 1 mol CuSO4 1 mol Cu 63.55 g Cu 1.274 g CuS O4 × × × = 0.5072 g Cu 159.62 g CuSO4 1 mol CuSO4 1 mol Cu Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be: Cancel units and calculate. actual yield percent yield = ( ) × 100 theoretical yield 0.392 g Cu = ( ) × 100 0.5072 g Cu = 77.3% Think about your result. Since the actual yield is slightly less than the theoretical yield, the percent yield is just under 100%.  EXERCISE 8.7.1 What is the percent yield of a reaction that produces 12.5 g of the Freon CF2Cl2 from 32.9 g of CCl4 and excess HF? CCl4 + 2 HF → CF2 Cl2 + 2 HCl Answer 48.3% SUMMARY Theoretical yield is calculated based on the stoichiometry of the chemical equation. The actual yield is experimentally determined. The percent yield is determined by calculating the ratio of actual yield to theoretical yield. 8.7: Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Masses of Reactants is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 8.7.3 https://chem.libretexts.org/@go/page/47507 8.8: Enthalpy Change is a Measure of the Heat Evolved or Absorbed When physical or chemical changes occur, they are generally accompanied by a transfer of energy. The law of conservation of energy states that in any physical or chemical process, energy is neither created nor destroyed. In other words, the entire energy in the universe is conserved. In order to better understand the energy changes taking place during a reaction, we need to define two parts of the universe, called the system and the surroundings. The system is the specific portion of matter in a given space that is being studied during an experiment or an observation. The surroundings are everything in the universe that is not part of the system. In practical terms for a laboratory chemist, the system is the particular chemicals being reacted, while the surroundings is the immediate vicinity within the room. During most processes, energy is exchanged between the system and the surroundings. If the system loses a certain amount of energy, that same amount of energy is gained by the surroundings. If the system gains a certain amount of energy, that energy is supplied by the surroundings. A chemical reaction or physical change is endothermic if heat is absorbed by the system from the surroundings. In the course of an endothermic process, the system gains heat from the surroundings and so the temperature of the surroundings decreases. The quantity of heat for a process is represented by the letter q. The sign of q for an endothermic process is positive because the system is gaining heat. A chemical reaction or physical change is exothermic if heat is released by the system into the surroundings. Because the surroundings are gaining heat from the system, the temperature of the surroundings increases. The sign of q for an exothermic process is negative because the system is losing heat. Figure 8.8.1 : (A) Endothermic reaction. (B) Exothermic reaction. Enthalpy Heat changes in chemical reactions are often measured in the laboratory under conditions in which the reacting system is open to the atmosphere. In that case, the system is at a constant pressure. Enthalpy (H) is the heat content of a system at constant pressure. Chemists routinely measure changes in enthalpy of chemical systems as reactants are converted into products. The heat that is absorbed or released by a reaction at constant pressure is the same as the enthalpy change, and is given the symbol ΔH. Unless otherwise specified, all reactions in this material are assumed to take place at constant pressure. The change in enthalpy of a reaction is a measure of the differences in enthalpy of the reactants and products. The enthalpy of a system is determined by the energies needed to break chemical bonds and the energies needed to form chemical bonds. Energy needs to be put into the system in order to break chemical bonds, as they do not come apart spontaneously in most cases. Bond formation to produce products will involve release of energy. The change in enthalpy shows the trade-offs made in these two processes. Does it take more energy to break bonds than that needed to form bonds? If so, the reaction is endothermic and the enthalpy change is positive. If more energy is produced in bond formation than that needed for bond breaking, the reaction is exothermic and the enthalpy is negative. Several factors influence the enthalpy of a system. Enthalpy is an extensive property, determined in part by the amount of material we work with. The state of reactants and products (solid, liquid, or gas) influences the enthalpy value for a system. The direction of the reaction affects the enthalpy value. A reaction that takes place in the opposite direction has the same numerical enthalpy value, but the opposite sign. Thermochemical Equation When methane gas is combusted, heat is released, making the reaction exothermic. Specifically, the combustion of 1 mol of methane releases 890.4 kilojoules of heat energy. This information can be shown as part of the balanced equation: CH (g) + 2 O (g) → CO (g) + 2 H O (l) + 890.4 kJ 4 2 2 2 8.8.1 https://chem.libretexts.org/@go/page/47509 The equation tells us that 1 mol of methane combines with 2 mol of oxygen to produce 1 mol of carbon dioxide and 2 mol of water. In the process, 890.4 kJ is released and so it is written as a product of the reaction. A thermochemical equation is a chemical equation that includes the enthalpy change of the reaction. The process in the above thermochemical equation can be shown visually in Figure 8.8.2. Figure 8.8.2 : (A) As reactants are converted to products in an exothermic reaction, enthalpy is released into the surroundings. The enthalpy change of the reaction is negative. (B) As reactants are converted to products in an endothermic reaction, enthalpy is absorbed from the surroundings. The enthalpy change of the reaction is positive. In the combustion of methane example, the enthalpy change is negative because heat is being released by the system. Therefore, the overall enthalpy of the system decreases. The heat of reaction is the enthalpy change for a chemical reaction. In the case above, the heat of reaction is −890.4 kJ. The thermochemical reaction can also be written in this way: CH (g) + 2 O (g) → CO (g) + 2 H O (l) ΔH = −890.4 kJ 4 2 2 2 Heats of reaction are typically measured in kilojoules. It is important to include the physical states of the reactants and products in a thermochemical equation as the value of the ΔH depends on those states. Endothermic reactions absorb energy from the surroundings as the reaction occurs. When 1 mol of calcium carbonate decomposes into 1 mol of calcium oxide and 1 mol of carbon dioxide, 177.8 kJ of heat is absorbed. The process is shown visually in Figure 8.8.2B. The thermochemical reaction is shown below. CaCO (s) + 177.8 kJ → CaO (s) + CO (g) 3 2 Because the heat is absorbed by the system, the 177.8 kJ is written as a reactant. The heat of reaction is positive for an endothermic reaction. CaCO (s) → CaO (s) + CO (g) ΔH = 177.8 kJ 3 2 The way in which a reaction is written influences the value of the enthalpy change for the reaction. Many reactions are reversible, meaning that the product(s) of the reaction are capable of combining and reforming the reactant(s). If a reaction is written in the reverse direction, the sign of the ΔH changes. For example, we can write an equation for the reaction of calcium oxide with carbon dioxide to form calcium carbonate. CaO (s) + CO (g) → CaCO (s) + 177.8 kJ 2 3 The reaction is exothermic and thus the sign of the enthalpy change is negative. CaO (s) + CO (g) → CaCO (s) ΔH = −177.8 kJ 2 3 Stoichiometric Calculations and Enthalpy Changes Chemistry problems that involve enthalpy changes can be solved by techniques similar to stoichiometry problems. Refer again to the combustion reaction of methane. Since the reaction of 1 mol of methane released 890.4 kJ, the reaction of 2 mol of methane 890, 4 kJ would release 2 × 890.4 kJ = 1781 kJ. The reaction of 0.5 mol of methane would release = 445.2 kJ. As with other 2 stoichiometry problems, the moles of a reactant or product can be linked to mass or volume.  Example 8.8.1 Sulfur dioxide gas reacts with oxygen to form sulfur trioxide in an exothermic reaction, according to the following thermochemical equation. 2 SO (g) + O (g) → 2 SO (g) + 198 kJ 2 2 3 8.8.2 https://chem.libretexts.org/@go/page/47509  Calculate the enthalpy change that occurs when 58.0 g of sulfur dioxide is reacted with excess oxygen. Solution Step 1: List the known quantities and plan the problem. Mass SO = 58.0 g 2 Molar mass SO = 64.07 g/mol 2 ΔH = −198 kJ for the reaction of 2 mol SO 2 Unknown ΔH =? kJ The calculation requires two steps. The mass of SO 2 is converted to moles. Then the moles of SO 2 is multiplied by the −198 kJ conversion factor of (. ) 2 mol SO 2 Step 2: Solve. 1 mol SO −198 kJ 2 ΔH = 58.0 g SO × × = 89.6 kJ 2 64.07 g SO 2 mol SO 2 2 Step 3: Think about your result. The mass of sulfur dioxide is slightly less than 1 mol. Since 198 kJ is released for every 2 mol of SO that reacts, the heat 2 released when about 1 mol reacts is one half of 198. The 89.6 kJ is slightly less than half of 198. The sign of ΔH is negative because the reaction is exothermic. 8.8: Enthalpy Change is a Measure of the Heat Evolved or Absorbed is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 8.8.3 https://chem.libretexts.org/@go/page/47509