chapter04_part1

# chapter04_part1 - Chapter 4: Quantities of Reactants and...

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Unformatted text preview: Chapter 4: Quantities of Reactants and Products 115 End-of-Chapter Solutions for Chapter 4 Summary Problem Answer: (a) 10.0 g CO (b) 8.27 g CO 2 (c) CO is limiting (d) Fe 2 O 3 is limiting (e) 94.7% Strategy and Explanation: Given the balanced equation for a reaction and the mass of one reactant, determine the mass of the other reactant, determine the mass of the product produced. Given a graph of product mass produced while varying the mass of one reactant, determine the limiting reactant under specific conditions. Given masses of all the reactants used to make a specific mass of product, determine the percent yield. (a) Use the molar mass of the reactant to find the moles. Then use the stoichiometry of the equation to determine the moles of the other reactant required. Then use the molar mass of it to find the grams. The balanced equation says: 3 mol CO reacts with 1 mol Fe 2 O 3 . 19.0 g Fe 2 O 3 " 1 mol Fe 2 O 3 159.688 g Fe 2 O 3 " 3 mol CO 1 mol Fe 2 O 3 " 28.0101 g CO 1 mol CO = 10.0 g CO (b) Use the molar mass of the reactant to find the moles. Then use the stoichiometry of the equation to determine the moles of the product produced. Then use the molar mass of it to find the grams. The balanced equation says: 3 mol CO 2 is produced by 1 mol Fe 2 O 3 . 10.0 g Fe 2 O 3 " 1 mol Fe 2 O 3 159.688 g Fe 2 O 3 " 3 mol CO 2 1 mol Fe 2 O 3 " 44.0095 g CO 2 1 mol CO 2 = 8.27 g CO 2 (c) When less than 10.0 grams of CO are available, CO is limiting. The graph shows that the yield is directly proportional to the amount of CO added. The calculation in (a) also shows that 10.0 g CO is required to completely consume 19.0 g Fe 2 O 3 . (d) When more than 10.0 grams of CO are available, Fe 2 O 3 is limiting. The graph shows that the yield is constant with the increase in the amount of CO added. The calculation in (a) also shows that only 10.0 g CO will be used to completely consume 19.0 g Fe 2 O 3 . (e) Use a metric conversion and the molar mass of the reactants to find the moles of the reactant substances present initially. Then use the stoichiometry of the equation to determine the moles of one of the products. Identify the limiting reactant from the reactant that produces the least number of products. From the limiting reactant quantity, determine the moles of the product, then use the molar mass of the product to get the grams and a metric conversion to get kilograms. The balanced equation says: 1 mol Fe 2 O 3 produces 2 mol Fe. 24.0 g Fe 2 O 3 " 1 mol Fe 2 O 3 159.6882 g Fe 2 O 3 " 2 mol Fe 1 mol Fe 2 O 3 = 0.301 mol Fe The balanced equation says: 3 mol CO produces 2 mol Fe. 20.0 g CO " 1 mol CO 28.0101 g CO " 2 mol Fe 3 mol CO = 0.476 mol Fe The number of Fe moles produced from Fe 2 O 3 is smaller (0.301 mol < 0.476 mol), so Fe 2 O 3 is the limiting reactant and CO is the excess reactant....
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## This note was uploaded on 12/17/2009 for the course CHE 129/130 taught by Professor Hanson during the Spring '09 term at SUNY Stony Brook.

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chapter04_part1 - Chapter 4: Quantities of Reactants and...

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