Sample Problem 510 Calculating the Amount of Gas Collected over Water PLAN The

# Sample problem 510 calculating the amount of gas

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Sample Problem 5.10 Calculating the Amount of Gas Collected over Water PROBLEM: Acetylene (C 2 H 2 ) is produced in the laboratory when calcium carbide (CaC 2 ) reacts with water: CaC 2 ( s ) + 2H 2 O( l ) C 2 H 2 ( 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 )
5-47 Sample Problem 5.10 multiply by M P total P of C 2 H 2 mass of C 2 H 2 use ideal gas law n of C 2 H 2 subtract P for H 2 O PLAN: SOLUTION: P C 2 H 2 = (738 - 21) torr = 717 torr 1 atm 760 torr = 0.943 atm P = 717 torr x 1 L 10 3 mL = 0.523 L V = 523 mL x T = 23°C + 273.15 K = 296 K
5-48 = 0.0203 mol 0.0203 mol x 26.04 g C 2 H 2 1 mol C 2 H 2 = 0.529 g C 2 H 2 SOLUTION: Sample Problem 5.10 0.943 atm 0.523 L x atm·L mol·K 0.0821 x 296 K n C 2 H 2 = RT PV =
5-49 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, T of gas A Amount (mol) of gas A Amount (mol) of gas B P , V, T of gas B
Sample Problem 5.11Using Gas Variables to Find Amounts of Reactants and Products IPROBLEM:What volume of H2gas at 765 torr and 225°C 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. M
5-51 SOLUTION: CuO( s ) + H 2 ( g ) Cu( s ) + H 2 O( g ) = 0.446 mol H 2 = 18.1 L H 2 Sample Problem 5.11 35.5 g CuO x 1 mol CuO 79.55 g CuO 1 mol H 2 1 mol CuO x 1 atm 760 torr = 1.01 atm P = 765 torr x T = 225°C + 273.15 K = 498 K V = P nRT 0.446 mol H 2 x x 498 K atm·L mol·K 0.0821 1.01 atm =
5-52 Sample Problem 5.12 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? PLAN: First we must write a balanced equation. Since the quantities of both reactants are given, we must next determine which reactant is limiting. We will use the ideal gas law to calculate the moles of Cl 2 present. SOLUTION: The balanced equation is: Cl 2 ( g ) + 2K( s ) → 2KCl( s ) For Cl 2 : P = 0.950 atm V = 5.25 L T = 293 K n = unknown
5-53 = 0.435 mol KCl 0.207 mol Cl 2 x 2 mol KCl 1 mol Cl 2 Sample Problem 5.12 0.50 atm 5.253 L x atm·L mol·K 0.0821 x 293 K n Cl 2 = RT PV = = 0.207 mol Cl 2 = 30.9 g KCl 0.435 mol KCl x 74.55 g KCl 1 mol KCl = 0.414 KCl 17.0 g K x 1 mol K 39.10 g K 2 mol KCl 2 mol K x For Cl 2 : For K: Cl 2 is the limiting reactant.
5-54 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.