Sample Problem 5.10Calculating the Amount of Gas Collected over WaterPROBLEM: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)
5-47Sample Problem 5.10multiply by MPtotalP of C2H2mass of C2H2use ideal gas lawnof C2H2subtract Pfor H2OPLAN:SOLUTION:PC2H2= (738 - 21) torr = 717 torr1 atm760 torr= 0.943 atmP = 717 torr x1 L103mL= 0.523 LV = 523 mL xT= 23°C + 273.15 K = 296 K
5-48= 0.0203 mol0.0203 mol x26.04 g C2H21 mol C2H2= 0.529 g C2H2SOLUTION:Sample Problem 5.100.943 atm0.523 Lxatm·Lmol·K0.0821x296 KnC2H2=RTPV=
5-49Figure 15.13The 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 StoichiometryP, V, Tof gas AAmount (mol)of gas AAmount (mol)of gas BP, V, Tof 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
5-51SOLUTION:CuO(s) + H2(g) →Cu(s) + H2O(g)= 0.446 mol H2= 18.1 L H2Sample Problem 5.1135.5 g CuO x1 mol CuO79.55 g CuO1 mol H21 mol CuOx1 atm760 torr= 1.01 atmP = 765 torr xT= 225°C + 273.15 K = 498 KV =PnRT0.446 mol H2xx 498 Katm·Lmol·K0.08211.01 atm=
5-52Sample Problem 5.12Using Gas Variables to Find Amounts of Reactants and Products IIPROBLEM: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 Cl2present.SOLUTION:The balanced equation is:Cl2(g) + 2K(s) → 2KCl(s)For Cl2:P= 0.950 atmV= 5.25 LT= 293 Kn= unknown
5-53= 0.435 mol KCl0.207 mol Cl2x2 mol KCl1 mol Cl2Sample Problem 5.120.50 atm5.253 Lxatm·Lmol·K0.0821x293 KnCl2=RTPV== 0.207 mol Cl2= 30.9 g KCl0.435 mol KCl x74.55 g KCl1 mol KCl= 0.414 KCl17.0 g K x1 mol K39.10 g K2 mol KCl2 mol KxFor Cl2:For K:Cl2is the limiting reactant.
5-54The Kinetic-Molecular Theory:A Model for Gas BehaviorPostulate 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.