General Chemistry I Modules 4-6.pdf.pdf

11 GENERAL CHEMISTRY 1 Quarter 1: Module 4-6 1  R epublicof the P hilippines D epartment of Education N a t i o n a l C a pi t a l Re g i o n Sc h o o l s D i v i s i o n O f f i c...

11 GENERAL CHEMISTRY 1 Quarter 1: Module 4-6 1  R epublicof the P hilippines D epartment of Education N a t i o n a l C a pi t a l Re g i o n Sc h o o l s D i v i s i o n O f f i c e o f La s Pi ñ a s C i t y DEVELOPMENT TEAM OF THE MODULE WRITERS: MADONNA L. MADRIDANO, Master Teacher I CHARISMA M. FERRER, Teacher III MARY ANN M. GUEVARRA, Teacher III CONSOLIDATOR: CHARISMA M. FERRER, Teacher III LANGUAGE EDITOR: CATHERINE P. DELOS REYES, Teacher I CONTENT MARITES T. TUDIO, Master Teacher I VALIDATORS: MARIEL B. BARAGENIO, Teacher III REX S. LAPID, Teacher III COVER PAGE AIRA MARI CON M. AUSTERO ILLUSTRATOR: TEAM LEADER: DR. RAQUEL M. AUSTERO Education Program Supervisor 2  Module 4 Limiting and Excess Reactant Most Essential Learning Competencies 1. Calculate percent yield and theoretical yield of the reaction (STEM_GC11MR-Ig- h-39) 2. Explain the concept of limiting reagent in a chemical reaction; identify the excess reagent(s) (STEM_GC11MR-Ig-h40) 3. Determine mass relationship in a chemical reaction (STEM_GC11MR-Ig-h42) What’s In Limiting and Excess Reactant The reactant in a chemical reaction that limits the amount of product that can be formed is called limiting reagent while the reactant that remain is the excess reagent. The reaction will stop when all the limiting reagent is consumed. The excess reagent remains because there is nothing with which it can react. Consider the figure below: No matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made. Likewise in chemistry, if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up. Steps in Identifying the Limiting Reactant To determine which is the limiting reactant, the amount of product that can be formed by each reactant must be calculated. The reactant that produces the less amount of product is the limiting reactant. 3  1. Balance the chemical equation. 2. Use stoichiometry to calculate the mass of the product that can be formed by each reactant. 3. The reactant that produces a lesser amount of product is the limiting reagent. 4. The reactant that produces a larger amount of product is the excess reagent. 5. To find the amount of excess reactant that remains after the reaction, subtract the mass of excess reagent that reacted from the given mass of the excess reactant. Example 1. A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) Solution. The equation is already balanced. Use stoichiometry to calculate how much product is produced by each reactant: Using 2.00 g NH3: 2.00 𝑔 𝑋𝑔 = 120 𝑔/𝑚𝑜𝑙 68.0 𝑔/𝑚𝑜𝑙 Xg = 3.53 g NO Using 4.00 g O2: 4.00 𝑔 𝑋𝑔 = 120 𝑔/𝑚𝑜𝑙 160 𝑔/𝑚𝑜𝑙 Xg = 3.00 g NO The reactant that produces the lesser amount of product is the oxygen. To find the amount of excess reactant that remains after the reaction, we must calculate how much of the excess reactant (NH3) actually did react with the limiting reactant (O2). 4.00 𝑔 𝑋𝑔 = 68.0 𝑔/𝑚𝑜𝑙 160 𝑔/𝑚𝑜𝑙 Xg = 1.70 g To find the amount of excess reactant that remains, subtract the amount that reacted from the given original amount. Excess reactant = 2.00 g – 1.70 g = 0.30 g 4 Example 2. Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2 Al + 3 Cl2 → 2 AlCl3 a. How many grams of aluminum chloride could be produced from 34.0 g of aluminum and 39.0 g of chlorine gas? b. Which is the limiting reagent? c. How much of the excess reactant remains after the reaction? Solution. The balanced equation is: 2 Al + 3 Cl2 → 2 AlCl3 Using 34.0 g Al: 34.0 𝑔 𝑋𝑔 = 264 𝑔/𝑚𝑜𝑙 54 𝑔/𝑚𝑜𝑙 Xg = 166 g AlCl3 Using 39.0 g Cl: 39.0 𝑔 𝑋𝑔 = 264 𝑔/𝑚𝑜𝑙 210 𝑔/𝑚𝑜𝑙 Xg = 49.0 g AlCl3 a. 49.0 g AlCl3 is produced. b. Cl2 is the limiting reactant. To find the amount of excess reactant that remains after the reaction, we must calculate how much of the excess reactant (Al) actually did react with the limiting reactant (Cl2). 39.0 𝑔 𝑋𝑔 = 54.0 𝑔/𝑚𝑜𝑙 210 𝑔/𝑚𝑜𝑙 Xg = 10.0 g To find the amount of excess reactant that remains, subtract the amount that reacted from the given original amount. Excess reactant = 34.0 g – 10.0 g = 24.0 g 5 Example 3. How many moles of NH3 are produced with 3.2 moles of N2 and 5.4 moles of H2? What is the limiting reactant? Which reactant is in excess and how much of the excess reactant remained after the reaction? N2 + 3 H2 → 2 NH3 Solution. The balanced equation is: N2 + 3 H2 → 2 NH3 Using 3.2 moles of N2: 3.2 𝑚𝑜𝑙 𝑋𝑚𝑜𝑙 = 1 2 Xmol = 6.4 mol NH3 Using 5.4 moles of H2: 5.4 𝑚𝑜𝑙 𝑋𝑚𝑜𝑙 = 3 2 Xg = 3.6 mol NH3 a. 3.6 mol NH3 is produced. b. H2 is the limiting reactant. c. N2 is the excess reactant. To find the amount of excess reactant that remains after the reaction, we must calculate how much of the excess reactant (N2) actually did react with the limiting reactant (H2). 𝑋𝑚𝑜𝑙 5.4 𝑚𝑜𝑙 = 1 3 Xmol = 1.8 mol N2 To find the amount of excess reactant that remains, subtract the amount that reacted from the given original amount. Excess reactant = 3.2 mol – 1.8 mol = 1.4 mol 6 Percent and Theoretical Yield When chemical reactions take place, they are almost never 100% complete. A reaction may not go to 100% due to not all the reactants becoming involved, impurities in the reactants, loss of product due to filtering, or just not getting it all out of the vessel or container. Percent yield is very important in the manufacturing of products. A lot of time and money is spent in 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 that may lead to negative effects on businesses. To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry which refers to theoretical yield. The actual yield is the actual product collected and measured or experimentally determined. The percent yield is a ratio of actual yield and theoretical yield which can be expressed as: Percent Yield = Actual Yield x 100 % Theoretical Yield Example 1 Calcium carbonate is decomposed by heating, as shown in the following equation: CaCO3(s) CaO(s) + CO2(g) a. What is the theoretical yield of this reaction if 25.8 g CaCO3 is heated to find 12.6 g CaO? b. What is the percent yield? Solution: Given: mass of CaCO3 = 25.8 g Molar mass CaCO3 = 100.1 g Molar mass CaO = 56.1 g Actual yield = 12.6 g CaO Theoretical yield = ? (Hint: use mass to mass calculation) 7 Convert 25.8 g CaCO3 g CaO (to cancel units in getting the percent yield) 25.8 g CaCO3 x 1 mol CaCO3 x 1 mol CaO x 56.1 g CaO = 14.46 g CaO 100.1 g CaCO3 1 mol CaCO3 1 mol CaO Percent Yield = Actual Yield x 100% Theoretical Yield % Yield = 12.6 g CaO x 100 % 14.46 g CaO = 0.8714 x 100 = 87.14 % Example 2 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, 30.0 g KClO3 is heated until it completely decomposes. What is the theoretical yield of oxygen? The experiment is performed and the oxygen gas is collected and its mass is found to be 14.9 g. What is the percent yield for the reaction? Solution: 30 g KClO3 x 1 mol KClO3 x 3 mol O2 x 32.0 g O2 = 11.75 g O2 122.55 g KClO3 2 mol KClO3 1 mol O2 8  What’s More Activity 1: Limiting and Excess Reactant Determine the limiting and excess reactant for the following problems. Calculate the amount of excess reactant that remains after the reaction. 1. Given the reaction: I2O5(g) + 5 CO(g) → 5 CO2(g) + I2(g) a. 80.0 grams of iodine(V) oxide, I2O5, reacts with 28.0 grams of carbon monoxide, CO. How many grams of iodine will be formed? b. What is the limiting reactant? Excess reactant? c. How much of the excess reactant remains after the reaction? 2. Zinc and sulfur react to form zinc sulfide according to the equation: Zn + S → ZnS a. If 25.0 g of zinc and 30.0 g of sulfur are mixed, which is the limiting reactant? b. How many grams of ZnS will be formed? c. How many grams of the excess reactant will remain after the reaction is over? 9  3. When MoO3 and Zn are heated together they react according to the equation: 3 Zn(s) + 2 MoO3(s) → Mo2O3(s) + 3 ZnO(s) a. What mass of ZnO is formed when 20.0 grams of MoO3 is reacted with 10.0 grams of Zn? b. What is the limiting reactant? Excess reactant? c. How much of the excess reactant remains after the reaction? 4. Given the balanced equation: 4 FeCl3 + 3 O2 → 2 Fe2O3 + 6 Cl2 a. How many moles of chlorine gas can be produced if 4.0 moles of FeCl3 react with 4.0 moles of O2? b. What is the limiting reactant? Excess reactant? c. How much of the excess reactant remains after the reaction? 5. Use the following BALANCED equation. 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O a. If 15 g of C2H6 react with 45 g of O2, how many moles of water will be produced? b. What is the limiting reactant? Excess reactant? c. How much of the excess reactant remains after the reaction? 10 Activity 2: Percent and Theoretical Yield Solve the following problems. Show complete solution. 1) A student adds 200.0g of C7H6O3 to an excess of C4H6O3, this produces C9H8O4 and C2H4O2. Calculate the percent yield if 231 g of aspirin (C9H8O4) is produced in an experiment. C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2 2) Calculate the percentage yield if 550.0 g of toluene (C7H8) added to an excess of nitric acid (HNO3) provides 305 g of the p-nitrotoluene (C7H7NO2) product in a lab experiment. C7H8 + HNO3 → C7H7NO2 + H2O 3) Aluminum reacts with an aqueous solution containing excess copper (II) sulfate. If 1.85 g Al reacts and the percentage yield of Cu is 56.6%, what mass of Cu is produced? 2 Al + 3 CuSO4 → 3 Cu + Al2(SO4)3 4) The combustion of methane (CH4) produces carbon dioxide and water. Assume that 2.0 mol of CH4 burned in the presence of excess air. What is the percentage yield if in an experiment the reaction produces 87.0 g of CO 2? CH4 + 2 O2 → CO2 + 2 H2O 5) 15.3 grams of Lithium is dropped into a solution containing excess copper (II) phosphate. When the reaction is completed, 52.5 grams of copper is formed. What is the percent yield? 6 Li + Cu3(PO4)2 → 2 Li3PO4 + 3 Cu 11 Activity 3: Concept Map You have already learned the basic concepts in stoichiometry. Make your own schematic diagram or flowchart on how to determine the limiting and excess reagent. Cite also at least five (5) applications related to percent yield concept. 12  What I Have Learned Directions: Give what is being asked. Write your answer on the line. 1. _______________________refers to the reactant that is used up first in the chemical reaction. 2. ________________________ pertains to the reactant that is present in quantities greater than what is needed by the reaction. 3. The use of ___________ is necessary to calculate the mass of the product that can be formed by each reactant. 4. The reactant that produces a lesser amount of product is called _________. 5. The reactant that produces a larger amount of product is called _________. 6. To identify the amount of excess reactant that remains after the reaction, subtract the _____________ that reacted from the given ____________. 7. __________________________is referred to as the ratio of actual yield/theoretical yield. What I Can Do The concept of limiting and excess reagents is concerned with the amount of products that will be formed when two or more reactants are mixed. This concept has various real-life applications. For example, in a manufacturing industry, the concept of limiting reactant is applied so that the process is efficient and with minimal waste as possible. Another example is in cooking. If you run out of a specific ingredient, you could apply the principles of stoichiometry to know exactly the amount of each ingredient you would use. After learning the concept of limiting reactant, think of a specific scenario/situation in your day to day living wherein you could apply the principles of stoichiometry and limiting reactant. It must also show the importance of limiting reactant. 13  Module 5 The Gas Phase Most Essential Learning Competencies 1. Define pressure and give the common units of pressure (STEM_GC11G-Ih-i-43) 2. Use the gas laws to determine pressure, volume, or temperature of a gas under certain conditions of change (STEM_GC11G-Ih-i-45) 3. Use the ideal gas equation to calculate pressure, volume, temperature, or number of moles of a gas (STEM_GC11G-Ih-i-46) 4. Use Dalton’s law of partial pressures to relate mole fraction and partial pressure of gases in a mixture (STEM_GC11DL-Ii-47) What’s In Gas Laws Gases behave differently from solids and liquids, so there are different ways for studying the behavior of gases. Gases have neither definite shape nor volume, unlike solids and liquids. They can fill the container completely. There are three variables in measuring gases, namely: pressure, volume, and temperature. Pressure It is the amount of force exerted per unit area. 1 standard atmosphere (atm) = 760 mm Hg = 760 torr = 29.92 inches Hg = 14.7 pounds/in2 (psi) = 101.3 kPa (SI unit is PASCAL) 14 Volume The common unit for volume is the liter (L), but the SI unit for volume is m 3. The equivalence of the liter in SI units is: 1L = 1000 m3 = 1 dm3 = 1000 cm3 (or mL) Temperature The common unit for temperature is degree Celsius (oC), but the SI unit is Kelvin (K). The relationship between the units is K = oC + 273 Gas Laws Boyle’s Law It states that the volume of a given amount of gas is inversely proportional to its pressure at constant temperature. In terms of a proportion: V α 1/ P (at constant amount and temperature) In terms of an equation: V = k/P (at constant amount and temperature) PV = k or P1 V1 = P2 V2 Example 1. A gas sample occupies a volume of 2.5 L at a pressure of 1.5 atm. What would be the volume of the gas if its pressure is reduced to 1.0 atm at the same temperature? 15 Solution P1 V1 = P2 V2 (1.5 atm) (2.5 L) = (1.0 atm) (V2) V2 = 3.8 L Example 2. The gas inside a balloon has a volume of 15.0 L at a pressure of 2.00 atm. Calculate the pressure of the gas if its volume is compressed to 10.0 L at the same temperature. Solution P1 V1 = P2 V2 (2.00 atm) (15.0 L) = (P2) (10.0 L) P2 = 3.00 atm Charles’s Law It states that the volume of a given amount of gas is directly proportional to its absolute temperature at constant pressure. In terms of a proportion: V α T (at constant amount and pressure) In terms of an equation: V = k T (at constant amount and pressure) V/T=k or V1 / T1 = V2 / T2 Example 1. At 30.0 oC, the volume of a sample of air was 5.80 L. What would be the volume of the air sample if it is heated to 60.0 oC at the same pressure? Solution V1 / T1 = V2 / T2 5.80 L/(30.0 + 273) = V2/(60.0 + 273) (5.80 L) (333 K) = 303 K (V2) V2 = 6.37 L 16 Example 2. A given amount of oxygen gas has a volume of 25.0 L at a temperature of 37.0 oC and a pressure of 1.00 atm. At what temperature would this gas occupy a volume of 22.0 L at a pressure of 1.00 atm? Solution V1 / T1 = V2 / T2 (25.0 L) / (37.0 + 273) = (22.0 L) / T2 (310 K) (22.0 L) = (25. 0 L) (T2) T2 = 273 K Gay-Lussac’s Law It states that the pressure is directly proportional to its absolute temperature at constant volume. In terms of a proportion: P α T (at constant volume) In terms of an equation: P = k T (at constant volume) P/T=k or P1 / T1 = P2 / T2 Example 1. Determine the new pressure when a constant volume of gas at 1.00 atm is heated from 20.0 oC to 30.0 oC. Solution P1 / T1 = P2 / T2 1.00 atm / 20.0 + 273 = P2/ 30.0 + 273 P2 = 1.03 atm Example 2. A gas has a pressure of 699.0 mm Hg at 40.0 oC. What is the temperature in oC at standard pressure? Solution P1 / T1 = P2 / T2 699.0 Mm Hg / 40.0 + 273 = 760.0 mm Hg/ T2 T2 = 340.3 K = 67.3 oC 17 Combined Gas Law The combined gas law combines the three gas laws: Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant. In terms of an equation: P = k T (at constant volume) PV/T = k or P1V1 / T1 = P2V2 / T2 Example 1. Find the volume of a gas at STP when 2.00 liters is collected at 745 mm Hg and 25.0 degrees Celsius. Solution P1V1 / T1 = P2V2 / T2 (745 mm Hg) (2.00 L) / (25.0+ 273) = (760 mm Hg) (V2) /(273 K) V2 = 1.80 L Example 2. A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29.0 °C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Solution P1V1 / T1 = P2V2 / T2 (0.800 atm) (0.180 L) / (29.0+ 273) = (3.20 atm) (0.090 L) /(T2) T2 = 604 K = 331 oC Avogadro’s Law It states that the volume of a gas at a given temperature pressure is directly proportional to the number of moles contained in the volume. In terms of a proportion: V α n (at constant temperature and pressure) In terms of an equation: V = k n (at constant temperature and pressure) V/n = k or V1 / n1 = V2 / n2 18 Example 1. 1.00 mole of a gas occupies a volume of 22.4 L gas at 0 oC and 1.0 atm. What would be the volume of 7.50 mol of the gas at the same temperature and pressure? Solution V1 / n1 = V2 / n2 22.4 L / 1.00 mole = V2 / 7.50 mole V2 = 168 L Example 2. A 3.2 mol of gas has a volume of 3.8 L. Find the new volume if the amount of gas is increased to 9.5 mol. Solution V1 / n1 = V2 / n2 3.8 L / 3.2 mole = V2 / 9.5 mole V2 = 11 L Ideal Gas Law The ideal gases obey the ideal gas law perfectly. This law states that the volume of a given amount of gas is directly proportional to the number of moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e. PV = nRT. The value of R can be calculated from the molar volume at 0oC and 1 atm (V=22.4 L). R = 0.0821 L atm/mol K Example 1. At what temperature will 0.654 moles of neon gas occupy 12.3 liters at 1.95 atmospheres? Solution PV = nRT T = PV/nR T = (1.95 atm) (12.3 L) / (0.654 mol) (0.0821 L atm/mol K) T = 447 K 19 Example 2. Determine the volume occupied by 2.34 grams of carbon dioxide gas at STP. Solution PV = nRT V = (2.34 g/44 g/mol) (0.0821 L atm/mol K) (273 K) / 1 atm V = 1.19 L Dalton’s Law of Partial Pressures It states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases: PT = P1 + P2 + P3 +…. Example 1. A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is the total pressure inside the container? Solution PT = P1 + P2 + P3 PT = 2.00 atm + 3.00 atm + 4.00 atm = 9.00 atm Example 2. A mixture of 2.00 moles of H2, 3.00 moles of NH3, 4.00 moles of CO2, and 5.00 moles of N2 exert a total pressure of 800.0 torr. What is the partial pressure of each gas? Solution PH2 = (2.00 moles/14.0 moles) (800.0 torr) = 114 torr PNH3 = (3.00 moles/14.0 moles) (800.0 torr) = 171 torr PCO2 = (4.00 moles/14.0 moles) (800.0 torr) = 229 torr PN2 = (5.00 moles/14.0 moles) (800.0 torr) = 286 torr 20  What’s More Activity 1: Question and Answer Directions: Answer the following questions concisely. 1. What is pressure and what causes it? _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 2. What are the common units of pressure? List them in order of smallest to largest unit. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ Activity 2: The Gas Laws: Problem Solving Directions: Solve the following problems. Show work and solution. Express final answers to the correct number of significant figures. 1. A sample of helium gas at 25.0°C is compressed from 200 cm3 to 0.240 cm3. Its pressure is now 3.00 cm Hg. What was the original pressure of the helium at constant temperature? 2. A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature? 21  3. The temperature of a sample of gas in a steel container at 30.0 kPa is increased from -100.0°C to 1.00 x 103 °C. What is the final pressure inside the tank? 4. How many moles of gas would be present in a gas trapped within a 100.0 mL vessel at 25.0°C at a pressure of 2.50 atmospheres? 5. Determine the volume occupied by 2.34 grams of Nitrogen gas (N2) at STP. 6. A 2.1 L flask contains 4.65 g of gas at 1.0 atm and 27 oC, what is the molar mass of the gas? 7. A gas mixture contains three components: 35 g CO, 3.8 g H2 and 12.3 g of O2. If the gases are at STP, what are the partial pressures of each gas? Activity 3: Challenge Directions: Solve the following problems. Show work and solution. Express final answers to the correct number of significant figures. 1. A gas occupies 55.0 Liters at a temperature of 85.0 oC. What will its volume be if it is cooled to 35.0 oC? 22 2. A gas originally at a pressure of 795 mm Hg and a volume of 510.0 mL is allowed to expand to 1.20 Liter. What is the new pressure of the gas? 3. Calculate the pressure of 2.60 Liters of a gas at 25.0 oC if it has a volume of 4.60 Liters at STP. 4. How many moles of gas will occupy a volume of 265.0 mL if 4.25 moles occupy a volume of 350.0 mL? 5. A mixture of gases contains oxygen, water vapor, carbon dioxide, and argon. If the partial pressure of oxygen is 3.8 atm, the partial pressure of the water vapor is 392 mmHg, the partial pressure of the carbon dioxide is 236 kPa, and the partial pressure of the argon is 148 mmHg, what is the total pressure? 6. A mixture of 45.0 g of oxygen and 45.0 g of helium has a total pressure of 0.990 atm. What is the partial pressure of each gas? 23  What I Have Learned Directions: Fill in the blanks with the correct answer. 1. _____________________is defined as the amount of force exerted per unit area. 2. Boyle’s Law states that __________________________________________. 3. Charles’s Law states that ________________________________________. 4. Gay-Lussac’s Law states that ____________________________________. 5. Combined Gas Law states that ___________________________________. 6. Avogadro’s Law states that ______________________________________. 7. Ideal Gas Law states that ________________________________________. 8. Dalton’s Law of Partial Pressures states that _______________________. What I Can Do The temperature has a great impact on the molecules of a gas. In a cooking gas (LPG) that we use at home which is composed of propane and butane, 1. suggest on what to do when temperature on the cooking area is high? 2. How can you apply the Gas laws in this situation? 24  Module 6 Gas Stoichiometry and Graham’s Law Most Essential Learning Competencies 1. Apply the principles of stoichiometry to determine the amounts (volume, number of moles, or mass) of gaseous reactants and products (STEM_GC11GS-Ii-j-48) 2. Relate the rate of gas effusion with molar mass (STEM_GC11KMT-Ij-50) What’s In Gas Stoichiometry Stoichiometry deals with the quantitative relationships between reactants and products in chemical reactions. We can calculate the amounts of reactants and products involved in chemical reactions. Previously, we have used masses for the amounts of reactants and products; in gaseous stoichiometry, we can express the amounts in gas volumes. Gas stoichiometry deals with gaseous substances given the volume or the volume is required in a chemical reaction. If volume, pressure, and temperature are known, we can use the ideal gas equation to determine the number of moles of gas. If moles of gas are calculated, we can solve for the volume of the gas at any given temperature and pressure. Example 1. How many grams of LiOH are needed to remove all the CO2 from the enclosed space of a spaceship whose volume is 2.4×105 L? The partial pressure of CO2 is 7.9×10-3 atm and the temperature is 296 K. Solution. We begin by using the ideal gas law to calculate the moles of CO2 needing removal; thus (7.9×10-3 atm)(2.4 ×105 L) n= (0.0821 L⋅atm mol⋅K) (296 K) = 78.0 mol CO2 Next, we use the reaction’s stoichiometry to determine the required moles of LiOH 78.0 mol CO2 × 2 mol LiOH × 23.95 g LiOH = 3.7×103 g LiOH 1 mol CO2 mol LiOH 25 Example 2. Propane, C3H8(g), is used in gas grills to provide the heat for cooking. What volume of O2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion. C3H8 + 5 O2 → 3 CO2 + 4 H2O Solution. The ratio of the volumes of C3H8 and O2 will be equal to the ratio of their coefficients in the balanced equation for the reaction. From the equation, we see that one volume of C3H8 will react with five volumes of O2: 2.7 L C3H8 x 5 L O2 = 13.5 L O2 1 L C3 H8 Graham’s Law of Diffusion Graham's law is a gas law which relates the rate of diffusion or effusion of a gas to its molar mass. Diffusion is the process of slowly mixing two gases together. Effusion is the process that occurs when a gas is permitted to escape its container through a small opening. Graham's law states that the rate at which a gas will effuse or diffuse is inversely proportional to the square root of the molar masses of the gas. This means light gasses effuse/diffuse quickly and heavier gases effuse/diffuse slowly. Thus, the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Example 1. If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster? Solution 1) Set rates and get atomic weights: RateA = He = x RateB = Ar = 1 The atomic weight of He = 4.00 The atomic weight of Ar = 40.0 2) Graham's Law is: RA / RB= √MMB / MMA 3) Substituting, we have: x / 1 = √40.0 / 4.00 x = 3.16 Helium escapes faster than Ar. It does so at 3.16 times the rate of the argon. 26  Example 2. What is the molecular weight of a gas which diffuses 1/50 as fast as hydrogen? Solution 1) Set rates and molecular weights: RateA = unknown gas = 1 RateB = H2 = 50 Remember, 1/50th of 50 is 1. The molecular weight of H2 = 2.0 The molecular weight of the other gas = x 2) By Graham's Law we have: 1 / 50 = √2.0 / x 0.0004 = 2.0 / x