# 5 44 sample problem 511 calculating the amount of gas

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and determining its pressure.
Sample Problem 5.11 Calculating the Amount of Gas Collected over Water
The
PROBLEM: Acetylene (C2H2) is produced in the laboratory when calcium carbide (CaC2) reacts with water: CaC2(s) + 2H2O (l) C2H2(g) + Ca(OH)2(A collected sample of acetylene has a total gas pressure of 738 torr and a volume of 523 mL. At the temperature of the gas (23ºC), the vapor pressure of water is 21 torr. How many grams of acetylene are collected? aq)
5-45 Sample Problem 5.11 multiply by MPtotalP of C2H2mass of C2H2use ideal gas law nof C2H2subtract Pfor H2O PLAN: SOLUTION: P C2H2= (738 - 21) torr = 717 torr 1 atm 760 torr = 0.943 atm P = 717 torr x 1 L 103mL = 0.523 L V = 523 mL x T= 23ºC + 273.15 K = 296 K
5-46 = 0.0203 mol 0.0203 mol x 26.04 g C2H21 mol C2H2= 0.529 g C2H2SOLUTION: Sample Problem 5.11 0.943 atm 0.523 L x atm·L mol·K 0.0821 x 296 K n C2H2= RT PV =
5-47 Figure 15.13 The relationships among the amount (mol, n) of gaseous reactant (or product) and the gas pressure (P), volume (V), and temperature (T). The Ideal Gas Law and Stoichiometry P, V, Tof gas AAmount (mol) of gas A Amount (mol) of gas B P, V, Tof gas B
Sample Problem 5.12 Using Gas Variables to Find Amounts of Reactants and Products I PROBLEM: What volume of H2gas at 765 torr and 225ºC is needed to reduce 35.5 g of copper(II) oxide to form pure copper and water? PLAN: Write a balanced equation. Convert the mass of copper(II) oxide to moles and find the moles of H2, using the mole ratio from the balanced equation. Calculate the corresponding volume of H2using the ideal gas law.
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5-49 SOLUTION: CuO (s) + H2(g) Cu (s) + H2O (g) = 0.446 mol H2= 18.1 L H2Sample Problem 5.12 35.5 g CuO x 1 mol CuO 79.55 g CuO 1 mol H21 mol CuO x 1 atm 760 torr = 1.01 atm P = 765 torr x T= 225ºC + 273.15 K = 498 KV =P nRT 0.446 mol H2x x 498 K atm·L mol·K 0.0821 1.01 atm =
Sample Problem 5.13 Using Gas Variables to Find Amounts of Reactants and Products II PROBLEM: What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g of potassium metal?
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5-51 = 0.414 mol KCl 0.207 mol Cl2x 2 mol KCl 1 mol Cl2Sample Problem 5.13 0.950 atm 5.25 L x atm·L mol·K 0.0821 x 293 K n Cl2= RT PV = = 0.207 mol Cl2= 30.9 g KCl 0.414 mol KCl x 74.55 g KCl 1 mol KCl = 0.435 KCl 17.0 g K x 1 mol K 39.10 g K 2 mol KCl 2 mol K x For Cl2: For K: Cl2is the limiting reactant because the given amount produces less KCl.
5-52 The Kinetic-Molecular Theory: A Model for Gas Behavior Postulate 1:Gas particles are tiny with large spaces between them. The volume of each particle is so small compared to the total volume of the gas that it is assumed to be zero.
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