Chemical Calculations Practice Questions PDF

EXPERIENCE fl Chemical Calculations GO ONLINE to Explore and Explain stoichiometric calculations....

EXPERIENCE fl Chemical Calculations GO ONLINE to Explore and Explain stoichiometric calculations. Mole Ratios A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles. In chemical calculations, mole ratios are used to convert between a given number of moles of reactant or product to moles of a different reactant or product. For example, mole ratios can be determined from the balanced chemical equation for ammonia production: N2(g) + 3H2(g) → 2NH3(g). Mole-Mole Graph The number of moles of NH3 is Mole Ratio The coefficients in the graphed as a function of the number of moles of N2 chemical equation can also be used to and H2. The coefficients in the chemical equation write the mole ratio between any of provide the proportionality between individual the reactants and/or products. reactants and/or products. Mole Ratio of Product:Reactants 1 mol N2 14 3 mol H2 12 2 10 = 1 2 mol NH3 8 Mol NH3 Copyright © Savvas Learning Company LLC. All Rights Reserved. 1 mol N2 6 4 2 = 2 3 2 mol NH3 0 0 2 4 6 8 10 12 14 3 mol H2 Mol reactant SEP Develop a Model What would the slopes of the lines be for a graph of the moles of product versus mole of each reactant for the reaction 2H2 + O2 → 2H2O? Ø Sample answer: The slope for the H2 line would be 1. The slope for the O2 line would be 2. 2 Chemical Calculations 259  Mole-Mole Calculations Mole ratios can be used in a mole-mole calculation, which is a conversion between a given number of moles of one reactant or product and moles of another reactant or product. If you’re given the number of moles of one of the substances, G, then you can use the appropriate mole ratio to calculate the moles of the wanted substance, W. To choose the correct mole ratio, make sure the moles of the wanted substance are in the numerator. Mole-Mole Flowchart Multiply the given number of moles by the appropriate mole ratio to calculate the moles of the wanted quantity. Given Wanted quantity quantity b mol W G × mol W a mol G Given Mole ratio from balanced Wanted quantity chemical equation quantity Applying the Mole-Mole Flowchart How many moles of O2 would you need to produce 25 mol H2O from the reaction of H 2 and O2? Write the balanced equation. 2H2 + O2 → 2H2O Develop a concept map for the a mol H2O × Mole b mol O2 ratio calculation. Given Wanted Copyright © Savvas Learning Company LLC. All Rights Reserved. quantity quantity Use the coefficients from the 2 mol H2O 1 mol O2 equation to write the mole ratios 1 mol O2 2 mol H2O that include H2O and O2. Choose the mole ratio that 1 mol O2 makes mol H2O cancel out and 25 × = 12.5 mol O2 calculate mol O2. SEP Use Mathematics How many moles of hydrogen gas would be required to produce 25 moles of H2O? Ø 25 moles of H2 260 Investigation 7 Stoichiometry  SAMPLE PROBLEM Calculating Moles of a Product How many moles of NH3 are produced when 0.60 mol of nitrogen react with hydrogen? ANALYZE List the knowns and unknown. Knowns Unknown moles of nitrogen = 0.60 mol N2 moles of ammonia = ? mol NH3 CALCULATE Solve for the unknown. N2 + H2 → NH3 (unbalanced) N2 + 3H2 → 2NH3 (balanced) 2 mol NH3 Remember that the mole ratio 1 mol N2 must have N2 on the bottom so that the mol N2 in the mole ratio will cancel with mol N2 in the known. 2 mol NH3 0.60 mol N2 × = 1.2 mol NH3 EVALUATE Does the result make sense? The ratio of 1.2 mol NH3 to 0.60 mol N2 is 2:1, as predicted by the balanced equation. Copyright © Savvas Learning Company LLC. All Rights Reserved. SEP Apply Mathematical Concepts This equation shows the formation of aluminum oxide, which is found on the surface of aluminum objects exposed to the air. Ø 4Al(s) + 3O2(g) → 2Al2O3(s) a. Write the six mole ratios that can be derived from this equation. 4 mol Al 3 mol O2 2 mol Al2O3 3 mol O 2 mol Al O , , 4 mol Al , 4 mol Al 3 mol O2 4 mol Al 2 mol Al2O3 , 2 mol Al 2O , 3 mol 2O 3 2 3 2 b. How many moles of aluminum are needed to form 3.7 mol Al2O3? 7.4 mol Al GO ONLINE for more practice problems. 2 Chemical Calculations 261  Mass-Mass Calculations Laboratory balances measure mass and cannot measure substances directly in moles. A mass-mass calculation is a conversion between a given mass of one reactant or product to the mass of another reactant or product. 1 mol G b mol W mass W mass of G × mol G × mol W × mass of W mass G a mol G 1 mol W Given Mole-mass Mole ratio from Mole-mass Wanted quantity relationship aG → bW relationship quantity Mass-to-Moles First, the mass Mole Ratio Then, the mole Moles-to-Mass Finally, the of the given component G is ratio is used to determine moles of W are converted converted to moles using the the number of moles of the to mass using the molar molar mass. wanted component W. mass of W. This method can be used to determine the mass of ammonia produced in the following reaction, if we start with 100 g of nitrogen. N2 + 3H2 → 2NH3 1 mol N2 2 mol NH3 17.0 g NH3 100 g N2 × 3.57 mol N2 × 7.14 mol NH3 × = 121 g NH3 28.0 N2 1 mol N2 Mass Ratio of Ammonia Reaction Two moles of ammonia are made when one mole of nitrogen reacts. However, the mass ratio is about 1.21 NH3:1 N2 because the molar masses are 17 g/mol and 28 g/mol, respectively. Mass Ratio of NH3:Nfl Copyright © Savvas Learning Company LLC. All Rights Reserved. 250 200 Mass NH3 (g) 150 You can make 121 g NH3 if you 100 have 100 g N2. 50 0 0 20 40 60 80 100 120 140 160 180 200 Mass N2 (g) SEP Develop a Model What are the mole ratio and the mass ratio for H2O to O2 in the reaction 2H2 + O2 → 2H2O? Ø 2 mol H2O:1 mol O2, 1.125 g H2O:1 g O2 262 Investigation 7 Stoichiometry  SAMPLE PROBLEM Calculating the Mass of a Product Calculate the number of grams of NH3 produced when 5.40 g H2 react with an excess of N2. The balanced equation is N2(g) + 3H2(g) → 2NH3(g). ANALYZE List the knowns and unknown. Knowns Unknown mass of hydrogen = 5.40 g H2 mass of ammonia = ? g NH3 1 mol H2 = 2.0 g H2 1 mol NH3 = 17.0 g NH3 CALCULATE Solve for the unknown. g H2 → mol H2 → mol NH3 → g NH3 determine the mass of ammonia. 1 mol H2 Convert the given mass to moles. 5.40 × 2.0 1 mol H2 2 mol NH3 Don’t forget to 5.40 × × 2.0 g H2 cancel the units at each step. 5.40 1 mol H2 17.0 g NH3 × × × = 31 g NH 3 2.0 g H2 1 Given Change Mole ratio Change quantity given unit moles to Copyright © Savvas Learning Company LLC. All Rights Reserved. to moles grams EVALUATE Does the result make sense? Multiple conversion factors make it difficult to evaluate the result, but the molar mass of NH3 is much greater than that of H2, which means the answer was likely to have a larger value than the given mass. SEP Apply Mathematical Concepts Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2). Ø CaC2(s) + 2H2O(l ) → C2H2(g) + Ca(OH)2(aq) How many grams of C2H2 are produced when water is added (in excess) to 5.00 g CaC2? 2.03 g C2H2 GO ONLINE for more practice problems. 2 Chemical Calculations 263  Volume-Volume Calculations So far, you’ve learned how to use the molar mass to convert between moles and mass, and how to use the mole ratio to complete the stoichiometric calculation. The mole can also be related to other quantities, such as volume. A volume-volume calculation is a conversion between a given volume of one reactant or product and the volume of another reactant or product. Volume-volume calculations only apply to gases at STP because the molar volume is used as a conversion factor. Volume of G 1 mol G Volume of W × mol G × mol W × 22.4 L W (liters) at STP 22.4 L G (liters) at STP Given Mole-volume Mole ratio from Mole-volume Wanted quantity relationship aG → bW relationship quantity Volume-to-Moles First, the Mole Ratio Then, the mole Moles-to-Volume Finally, volume of the given component ratio is used to determine the moles of W are converted G is converted to moles using the number of moles of the to volume using the molar the molar volume. wanted component W. volume of W. 2CO(g) + O2(g) → 2CO2(g) 1 mol O2 2 mol CO2 22.4 L CO2 1 L O2 × mol O2 × mol CO2 × = 2 L CO2 22.4 L O2 1 mol O2 1 mol CO2 Combustion of Carbon Monoxide To Volume COfl vs. CO and Ofl make 2 L of carbon dioxide, you need 1 L of oxygen. The volume ratio is 2 L CO2:1 L 20 Volume CO2 (L) O2. Because of Avogadro’s law, the mole 15 ratio is also 2 mol CO2:1 mol O2. Copyright © Savvas Learning Company LLC. All Rights Reserved. 10 O2 The slope of the line for O 5 2 CO is 2 because the volume 0 ratio is 2 L CO2: 1 L O2. 0 5 10 15 20 Volume reactants (L) The slope of the line for CO is 1 because the volume ratio is 1 L CO 2: 1 L CO. SEP Develop a Model What are the slopes of the lines on a graph of the volume of the product versus volume of the reactants for the reaction represented by 2NO(g) + O2(g) → 2NO2(g)? Ø 1 for NO2 versus NO, and 2 for NO2 versus O2 264 Investigation 7 Stoichiometry  SAMPLE PROBLEM Calculating the Volume of a Product Nitrogen monoxide (NO) and oxygen gas combine to form the brown gas nitrogen dioxide (NO2), which contributes to photochemical smog. How many liters of NO2 are produced when 34 L O2 react with an excess of NO at STP? The balanced equation is 2NO(g) + O2(g) → 2NO2(g). ANALYZE List the knowns and unknown. Knowns Unknown volume of oxygen = 34 L O2 volume of nitrogen dioxide = ? L NO2 1 mol O2 = 22.4 L O2 (at STP) 1 mol NO2 = 22.4 L NO2 (at STP) CALCULATE Solve for the unknown. L O2 → mol O2 → mol NO2 → L NO2 1 mol O2 34 × 2 mol NO2 34 × × 34 × 2 mol NO2 22.4 L NO2 = 68 L NO2 Copyright © Savvas Learning Company LLC. All Rights Reserved. Finish by using the molar volume × 1 mol O2 × 1 Given Change Mole ratio Change to quantity to moles liters EVALUATE Does the result make sense? Because 2 mol NO2 are produced for each 1 mol O2 that reacts, the volume of NO2 should be twice the given volume of O2. The answer should have two significant figures. SEP Apply Mathematical Concepts How many liters of oxygen are required to burn 3.86 L of carbon monoxide? Ø 2CO(g) + O2(g) → 2CO2(g) 1.93 L O2 GO ONLINE for more practice problems. 2 Chemical Calculations 265  A Roadmap for Solving Stoichiometric Problems Mass can be measured easily for solids and liquids, and volume can be measured easily for gases. What happens when a chemical reaction involves a combination of solids and gases? Combining the processes discussed so far provides a general framework that you can use to complete mass- volume, particle-mass, and volume-particle calculations. Mass-Volume The dashed arrows show the process for calculating the volume of a gas needed in a reaction when given the mass of another substance. moles of G moles of W mass of G × 1 mol G × mass W = mass of W mass G 1 mol W b mol W mol G × a mol G mol W volume of G × 1 mol G × 22.4 L W = volume of W at STP 22.4 L G Mole ratio from 1 mol W at STP balanced equation representative aG → bW 6.02 × 1023 representative 1 mol G = × × particles of G 6.02 × 1023 1 mol W particles of W Given quantity Wanted quantity Volume-Mass The dotted arrows show the process for calculating the mass of a substance needed in a reaction when given the volume of a gas. Copyright © Savvas Learning Company LLC. All Rights Reserved. SEP Computational Thinking Sketch a flowchart for determining the number of representative particles of W needed if given the mass of G. Ø 6.02 × 1023 particles W Mass of G × 1 mol G mol G × b mol W mol W × mass G a mol G 1 mol W = particles W 266 Investigation 7 Stoichiometry  SAMPLE PROBLEM Calculating Molecules of a Product How many molecules of oxygen are produced when 29.2 g of water are decomposed by electrolysis according to this balanced equation? 2H2O(l ) electricity 2H2(g) + O2(g) ANALYZE List the knowns and unknown. Knowns Unknown molecules of oxygen = mass of water = 29.2 g H2O ? molecules of O2 1 mol H2O = 18.0 g H2O 1 mol O2 = 6.02 × 1023 molecules O2 CALCULATE Solve for the unknown. g H2O → mol H2O → mol O2 → molecules O2 Convert the given mass 1 mol H2O 29.2 × 18.0 g H2O 1 mol O2 29.2 × × 18.0 g H2O 6.02 × 1023 molecules O2 29.2 × 18.0 g H O × mol H O × 2 2 Copyright © Savvas Learning Company LLC. All Rights Reserved. Given Change to Mole ratio Change to quantity moles molecules = 4.88 × 1023 molecules O2 EVALUATE Does the result make sense? The given mass of water should produce a little less than 1 mole of oxygen, which is a little less than Avogadro’s number of molecules. The answer has the right number of significant figures. SEP Apply Mathematical Concepts How many molecules of oxygen are produced when 6.54 g of potassium chlorate (KClO3) decompose? Ø 2KClO3(s) → 2KCl(s) + 3O2(g) 4.82 × 1022 molecules O2 GO ONLINE for more practice problems. 2 Chemical Calculations 267 Revisit GO ONLINE to Elaborate on and Evaluate your knowledge of stoichiometric calculations by PHENOMENON completing the peer review and writing activities. In the CER worksheet you completed at the beginning of the investigation, you drafted an explanation for why a recipe can fail. With a partner, reevaluate your arguments. SEP Use Mathematics When dough bakes in the oven, baking soda, also known as sodium bicarbonate (NaHCO3), decomposes into sodium carbonate (Na2CO3), water, and carbon dioxide. If your recipe calls for 2.0 grams of baking soda, what mass of carbon dioxide will be released during the reaction? Write a balanced chemical equation and use the molar masses in the table. Ø Molar Masses of Reactants and Products Substance Molar mass (g/mol) NaHCO3 84 Na2CO3 106 H2O 18 CO2 44 2NaHCO3 → Na2CO3 + H2O + CO2; 44 g = 0.52 g CO2 84 g Copyright © Savvas Learning Company LLC. All Rights Reserved. 268 Investigation 7 Stoichiometry  Limiting Reagent and Percent Yield GO ONLINE to Explore and Explain limiting reagents and reaction yields. Limiting Ingredients In any recipe, an insufficient amount of any of the ingredients will limit the amount of product you can make. The amount of product is determined by the amount of the limiting ingredient. No matter how much of the other ingredients you have, you can only make as much product as allowed by the limiting ingredient. For example, the limiting ingredient determines how many tacos you can make, even if you have more of every other ingredient. You could make more than two tacos with these ingredients. They are excess ingredients. Cilantro Salsa Lime Shredded cabbage Grilled fish Copyright © Savvas Learning Company LLC. All Rights Reserved. If you have two Tortillas are the tortillas, you limiting ingredient. can only make two tacos. Tortilla warmer Fish tacos Limiting and Excess Ingredients Combining a tortilla, fish, cabbage, cilantro, salsa, and a splash of lime produces a delicious taco. If you use up the tortillas before the other ingredients, they are the limiting ingredient. SEP Use a Model Write a balanced equation for the production of sausage sandwiches. Assume that sausages only come in packs of five and buns in packs of eight. How many sandwiches can you make if you have one pack of each ingredient? Ø 1 pack of 5 sausages + 1 pack of 8 buns → 5 sandwiches + 3 buns; 5 sandwiches 3 Limiting Reagent and Percent Yield 269  Limiting and Excess Reagents In a chemical reaction, an insufficient amount of any of the reactants will limit the amount of product that forms. The reactant that determines the amount of product that can be formed is called the limiting reagent. The reaction stops after the limiting reagent is used up, even though some amount of the other reactant(s) remains. Any reactant that is not used up in a reaction is called the excess reagent. Reactants Products Balanced Reactants The chemical equation for the N2(g) + 3H2(g) 2NH3(g) production of ammonia is a precise “recipe” that calls for three molecules of hydrogen (H2) for every one molecule of nitrogen (N2). The product is two + molecules of ammonia (NH3), with no leftover reagents. 1 molecule N2 + 3 molecules H2 2 molecules NH3 Unbalanced Reactants Before Reaction After Reaction In this particular experiment, H2 is the limiting reagent and N2 is the excess reagent. After all of the H2 is used up to make NH3, there is still + + some N2 left over. Copyright © Savvas Learning Company LLC. All Rights Reserved. 2 molecules N2 + 3 molecules H2 2 molecules NH3 + 1 molecule N2 SEP Interpret Data Varying masses of sodium metal 7 react with a fixed mass of chlorine. The graph shows the masses of sodium used and sodium chloride 6 produced. Explain the general shape of the graph Mass of NaCl (g) 5 and identify the limiting and excess reagents. Ø 4 Sample answer: The mass of NaCl is 3 proportional to the mass of Na, until there is a 2 1 flat line when there is not enough Cl to react 0 2 3 4 0 1 with all the Na. Beyond that point, Cl is the Mass of Na (g) limiting reagent, and Na is the excess reagent. 270 Investigation 7 Stoichiometry  Mass of Products and Reactants Often in stoichiometric problems, the given quantities of reactants are expressed in units other than moles, such as mass. The amount of each reactant first has to be converted from mass to moles before the mole ratio is applied to determine the limiting reagent. Then, the limiting reagent can be used to determine the mass of the products. Determining the Limiting Reagent The mole ratio is used to determine the number of moles of one of the reactants needed to complete the reaction. If you do not have enough, then this is the limiting reagent. aR1 + bR2 → cW How many moles Given Mole-mass are needed for the quantities relationship Mole ratio complete reaction? 1 mol R1 b mol R2 mass of R1 × mol R1 × mol R2 needed mass R1 a mol R1 The mass of the reactants is given. 1 mol R2 mass of R2 × mol R2 mass R2 Convert to moles. mol R2 < mol R2 needed R2 is limiting mol R2 > mol R2 needed R2 is excess Determining Mass from the Limiting Reagent The limiting reagent limits the reaction. It is used in a mole-mass calculation with the mole ratio to determine the mass of the product. Mole-mass Quantity Copyright © Savvas Learning Company LLC. All Rights Reserved. Mole ratio relationship wanted mol Rlimit a mol R mol W × mass W = mass of W limit 1 mol W SEP Analyze Data The table shows the masses of Na(g) + Clfl(g) Reaction Data sodium used and sodium chloride produced when varying masses of sodium react with a fixed mass Mass Na (g) Mass NaCl (g) of chlorine. From the data, estimate the total mass 1 2.5 of chlorine that was available for each trial. Ø 2 5 2.5 6.5 Sample answer: about 3.9 g Cl2 4 6.5 3 Limiting Reagent and Percent Yield 271  PROBLEM Determining the Limiting Reagent Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: 2Cu(s) + S(s) → Cu2S(s). What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? ANALYZE List the knowns and unknown. Knowns Unknown mass of copper = 80.0 g Cu limiting reagent = ? mass of sulfur = 25.0 g S CALCULATE Solve for the unknown. Choose one of the reactants and convert 1 mol Cu 80.0 × 63.5 = 1.26 mol Cu from mass to moles. Convert the other reactant from mass 1 mol S 25.0 × = 0.779 mol S to moles. 32.1 g S 1.26 × 1 mol S = 0.630 mol S Given Mole ratio Needed quantity amount 0.630 mol S (amount needed) < 0.779 mol S (given amount). Sulfur is in excess. Therefore, copper is the limiting reagent. Copyright © Savvas Learning Company LLC. All Rights Reserved. If you used the actual number of moles of S to find the amount of copper needed, then you would still identify copper as the limiting reagent. EVALUATE Does the result make sense? Since the ratio of the given moles of Cu to moles of S was less than the ratio (2:1) from the balanced equation, copper should be the limiting reagent. SEP Apply Mathematical Concepts The equation for the combustion of ethene (C2H4) is C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g). If 2.70 mol C2H4 reacts with 6.30 mol O2, identify the limiting reagent. Ø O2 is the limiting reagent. GO ONLINE for more practice problems. 272 Investigation 7 Stoichiometry  PROBLEM Using the Limiting Reagent to Find the Quantity of a Product What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu react with 25.0 g S? 2Cu(s) + S(s) → Cu2S(s) ANALYZE List the knowns and unknown. Knowns Unknown limiting reagent = 1.26 mol Cu yield = ? g Cu2S 1 mol Cu2S = 159.1 g Cu2S CALCULATE Solve for the unknown. mol Cu → mol Cu2S → g Cu2S determine the mass of product. Use the mole ratio from 1 mol Cu2S the balanced equation. 1.26 × 2 Finish by converting from moles 159.1 g Cu2S 1.26 × × = 1.00 × 102 g Cu 2S EVALUATE Does the result make sense? Copper is the limiting reagent. The maximum number of grams of Cu2S Copyright © Savvas Learning Company LLC. All Rights Reserved. produced should be more than the amount of Cu that reacted because Cu and S combine. However, the mass of Cu2S produced should be less than the total mass of the reactants (105.0 g) because S was in excess. SEP Apply Mathematical Concepts The incomplete combustion of ethene is given by the equation C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g). Ø If 2.70 mol C2H4 is reacted with 6.30 mol O2, a. identify the limiting reagent. C2H4 is the limiting reagent b. calculate the moles of water produced. 5.40 mol H2O GO ONLINE for more practice problems. 3 Limiting Reagent and Percent Yield 273  Percent Yield Success Stats In the average professional hockey game, a team will make thirty shots at the goal. Theoretically, every shot could result in a score. However, this outcome does not occur. The average professional team scores fewer than two goals per game. The success percentage is the ratio of the actual score to the theoretical score expressed as a percent: theoretical score Percent Success A hockey team may score on less than 7% of the Improving the Score To improve, the team could shots that they make on the goal. either try to make more shots on the goal during a game or increase their percent success by making 35 more of the shots they attempt. 30 25 Shots on Goal 20 15 10 5 0 Copyright © Savvas Learning Company LLC. All Rights Reserved. SEP Define a Problem Grades on tests can also be thought of as a success statistic. How would you define the actual score and theoretical score on an exam, and how would you calculate the percent success? Ø Sample answer: On a test, getting all the answers correct would be the theoretical score. The actual score would be how many you got correct. The percent success would be the number of questions you got correct divided by the number of total questions and multiplied by 100. 274 Investigation 7 Stoichiometry  Percent Yield in Reactions When a balanced chemical equation is used to calculate the amount of product of a reaction, the calculated value represents the theoretical yield. It is the maximum amount of product that can be formed. The actual amount of product made during a real reaction is called the actual yield. You can measure the mass of the product of a laboratory experiment to find the actual yield. The percent yield is the ratio of the actual to the theoretical yield expressed as a percent, and represents the efficiency of a reaction: Percent Yield = Before Reaction After Reaction Theoretical Yield The balanced N2(g) + 3H2(g) 2NH3(g) chemical equation for the formation of ammonia can be used to determine the theoretical yield. Theoretical yield assumes that all of the reactants are used to make the product. + Side Reactions and Actual Yield Hydrogen reacts Often, side reactions reduce the yield with oxygen in a of a reaction. In this case, only 50% of side reaction. the possible ammonia was produced because oxygen caused a side + reaction. The oxygen may have been introduced from an impure gas source. + Copyright © Savvas Learning Company LLC. All Rights Reserved. Before Reaction After Reaction 26 SEP Construct an Explanation Imagine you are a chemical engineer at a chemical plant. One day the air conditioning in the plant stops working, resulting in high humidity. The percent yield for a chemical you produce drops significantly. Construct an explanation for the decrease in yield. Ø Sample answer: Higher humidity means more water vapor in the air, which could increase the amount of reactant used up in side-reactions. This would lower the amount of reactant available for the desired chemical reaction and lower the yield. 3 Limiting Reagent and Percent Yield 275  PROBLEM Calculating the Theoretical Yield Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is ∆ CaCO3(s) ⟶ CaO(s) + CO2(g) What is the theoretical yield of CaO if 24.8 g CaCO3 is heated? ANALYZE List the knowns and unknown. Knowns Unknown mass of calcium carbonate = 24.8 g CaCO3 theoretical yield = ? g CaO 1 mol CaO = 56.1 g CaO CALCULATE Solve for the unknown. Theoretical yield is a mass-mass calculation. Identify the steps to determine g CaCO3 → mol CaCO3 → mol CaO → g CaO Convert the given mass of the 1 mol CaCO3 24.8 × reactant to moles of the reactant. 100.1 g CaCO3 1 mol CaCO3 1 mol CaO 24.8 × 100.1 g CaCO × 3 1 1 mol CaCO3 1 mol CaO 56.1 g CaO Finish by converting from moles 24.8 × 100.1 g CaCO × 1 mol CaCO × Copyright © Savvas Learning Company LLC. All Rights Reserved. 3 3 = 13.9 g CaO EVALUATE Does the result make sense? The mole ratio of CaO to CaCO3 is 1:1. The ratio of their masses in the reaction should be the same as the ratio of their molar masses, which is slightly greater than 1:2. The result of the calculations shows that the mass of CaO is slightly greater than half the mass of CaCO3. SEP Apply Mathematical Concepts When 84.8 g of iron(III) oxide react with an excess of carbon monoxide, iron is produced. What is the theoretical yield of iron? Ø Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) 59.3 g Fe GO ONLINE for more practice problems. 276 Investigation 7 Stoichiometry  PROBLEM Calculating the Percent Yield What is the percent yield if 13.1 g CaO are actually produced when 24.8 g CaCO3 are heated? ∆ CaCO3(s) ⟶ CaO(s) + CO2(g) ANALYZE List the knowns and unknown. Knowns Unknown actual yield = 13.1 g CaO percent yield = ? % theoretical yield = 13.9 g CaO CALCULATE Solve for the unknown. actual yield Substitute the values for actual yield and percent yield = × 100 theoretical yield 13.1 g CaO percent yield = × 100 = 94.2% 13.9 g CaO EVALUATE Does the result make sense? In this example, the actual yield is slightly less than the theoretical yield. Therefore, the percent yield should be slightly less than 100 percent. SEP Apply Mathematical Concepts If 50.0 g of silicon dioxide are heated with an excess of carbon, 27.9 g of silicon carbide are produced. Ø ∆ SiO2(s) + 3C(s) ⟶ SiC(s) + 2CO(g) What is the percent yield of this reaction? Copyright © Savvas Learning Company LLC. All Rights Reserved. 1 mol SiO2 1 mol SiC 40.097 g = 33.3 g SiC 50 g SiO2 × × × 60.084 g 1 mol SiO2 1 mol SiC 27.9 g SiC × 100 = 83.5% 33.3 g SiC SEP Apply Mathematical Concepts If 15.0 g of nitrogen react with 15.0 g of hydrogen, 10.5 g of ammonia are produced. What is the theoretical yield and percent yield of this reaction? Theoretical yield = 18.2 g ammonia; percent yield = 57.7% GO ONLINE for more practice problems. 3 Limiting Reagent and Percent Yield 277 Revisit GO ONLINE to Elaborate on and Evaluate your knowledge of percent yield, actual yield, and theoretical yield by completing the discussion and engineering design activities. In the CER worksheet you completed at the beginning of the investigation, you drafted an explanation for why a recipe can fail. With a partner, reevaluate your arguments. SEP Analyze and Interpret Data A simple bread recipe calls for 400 g of flour, 8 g of salt, 1 g of yeast, and 0.3 L of water. The recipe produces 1 loaf of bread. The data table shows the amounts of each ingredient you have. Identify the limiting reagent. What is the theoretical yield, assuming you could make partial loaves? How many whole loaves of bread can you actually make? What is the percent yield? How much of each ingredient do you have left over? Ø Available Ingredients Ingredients Amount (g) Flour 1,350 Salt 450 Yeast 7 Water unlimited Flour is the limiting ingredient. The theoretical yield is 3.38 loaves. The actual yield is 3 loaves. The percent yield is 88.8%. I will have left over 150 g of flour, Copyright © Savvas Learning Company LLC. All Rights Reserved. 426 g of salt, 4 g of yeast, and an unlimited amount of water. 278 Investigation 7 Stoichiometry