Gas Laws Lecture Examples Answer Key

Chapter 10 — Gas Laws Lecture Examples Answer Key Ex 1 Some nitrogen gas is in a 2.00-L tank at a pressure of 3.00 atm. The tank is connected to a 5.00-L tank that is completely empty (evacuated), and a valve connects the two tanks., If the valve is opened, determine the total pressure in this two-tank system after the nitrogen stops flowing. No temperature change occurs in the process. P1V1 = P2V2 P2 = P1V1/V2 = (3.00 atm)(2.00 L)/(7.00 L) = 0.857 atm Ex 2 A certain mass of C3H6 reacts completely with an excess of ammonia and oxygen to give 38.2 L of C3H3N at 75°C and P = 1 atm. If the same mass of C3H6 reacts completely at 350°C and P = 1 atm, what volume of C3H3N can be collected at the higher temperature? 2 C3H6(g) + 2 NH3(g) + 3 O2(g) → 2 C3H3N(g) + 6 H2O(g) V1/T1 = V2/T2 V2 = V1T2/T1 = (38.2 L)(623K)/(348K) = 68.4 L Ex 3 A gas is in a container with a movable piston under a pressure of 6.4 atm. The volume of the container is increased by a factor of 4.5 by moving the piston. The absolute temperature is simultaneously doubled. Calculate the pressure of the gas after these changes. P1 = 6.4 atm P2 = ?? V1 V2 = 4.5 V1 T1 T2 = 2 T1 P2 = P1V1T2/V2T1 = (6.4 atm)V1(2T1)/(4.5V1)T1 = 2.8 atm Ex 4 At what temperature (in K) does 29.8 g of O2 gas have a pressure of 2.00 atm in a 10.0-liter tank? n = (29.8 g)(1 mol / 31.9988 g) = 0.9313 mol T = PV/nR = (2.00 atm)(10.0 L) / (0.9313 mol)(0.08206 L·atm/mol·K) = 262K Ex 5 Calculate the density of gaseous SF6 at a temperature of 27°C and a pressure of 0.873 atm. 1 PV=nRT density = d = m/V n/V = P/RT m/V = PM/RT = (0.873 atm)(146.0564 g/mol)/(0.08206 L·atm/mol·K)(300K) = 5.18 g / L Ex 6 The empirical formula of a gaseous fluorocarbon is CF2. If 1.55 g of this compound occupies 0.174 L at STP determine the molecular formula of this compound. M = (m/V)RT/P = (1.55 g/0.174 L)(0.08206 L·atm/mol·K)(273K) / (1 atm) = 199.56 g / mol M CF2 = 50.0078 g / mol 199.56 / 50.0078 = 4 Molecular formula = C4F8 Ex 7 An important process for the production of acrylonitrile (C3H3N) is given by the following reaction: 2 C3H6(g) + 2 NH3(g) + 3 O2(g) → 2 C3H3N(g) + 6 H2O(g) A 150.0-L reactor is charged to the following partial pressures at 25°C: PC3H6 = 4.93 atm, PNH3 = 7.90 atm, PO2 = 14.8 atm What mass of acrylonitrile can be produced from this mixture? nC3H6 = (4.93 atm)(150.0 L)/R(298K) = 30.24 mol C3H6 nacrylonitrile = (30.24 mol C3H6)(2 mol acrylonitrile / 2 mol C3H6) = 30.24 mol nNH3 = (7.90 atm)(150.0L)/R(298K) = 48.46 mol NH3 nacrylonitrile = (48.46 mol NH3)(2 mol acrylonitrile / 2 mol NH3) = 48.46 mol nO2 = (14.8 atm)(150.0 L)/R(298K) = 90.78 mol O2 nacrylonitrile = (90.78 mol O2)(2 mol acrylonitrile / 3 mol O2) = 60.52 mol C3H6 is the limiting reactant mass acrylonitrile = (30.24 mol)(42.0804 g / 1 mol) = 1.27 kg Ex 8 A 0.326-g sample of XH2 reacts with water according to the equation 2 XH2(s) + 2 H2O(l) → + X(OH)2(s) 2 H2(g) 3 The hydrogen evolved is found to have a volume when dry of 375 cm at 21°C and 1.00 atm. What is the atomic mass of X? nH2 = (1.00 atm)(0.375 L)/R(294K) = 0.01554 mol nXH2 = (0.01554 mol H2)(1 mol XH2 / 2 mol H2) = 0.00777 mol XH2 molar mass XH2 = 0.326 g / 0.00777 mol = 41.96 g / mol molar mass of X = 41.96 g / mol – 2(1.0079 g / mol) = 39.94 g / mol Ex 9 Acetylene (C2H2) can be made by allowing calcium carbide to react with water. CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(aq) You place 2.65 g of CaC2 in excess water and collect the acetylene over water. (a) Calculate the theoretical yield of acetylene (in grams). mol CaC2 = (2.65 g)(1 mol / 64.10 g) = 0.04134 mol CaC2 mol C2H2 = (0.04134 mol)(1 mol C2H2 / 1 mol CaC2) = 0.04134 mol C2H2 mass C2H2 = (0.04134 mol)(26.0378 g / 1 mol) = 1.08 g (b) Theoretical yield After reaction, the volume of the acetylene and water vapor is 795 mL at 25.0°C and a pressure of 0.967 atm. Calculate the percent yield of acetylene. The vapor pressure of water at 25.0°C is 0.0313 atm. PC2H2 = 0.967 atm – 0.0313 atm = 0.9357 atm nC2H2 = PV/RT = (0.9357 atm)(0.795 L)/R(298K) = 0.03042 mol m = (0.03042 mol)(26.0378 g / 1 mol) = 0.792 g Actual yield % yield = (0.792 g / 1.08 g) × 100% = 73.3% Ex 10 The partial pressure of O2(g) in a mixture of O2(g) and H2(g) is 2.54 atm. The total pressure of the mixture is 3.92 atm. Determine the partial pressure of H2(g), and the mole fraction of each gas in the mixture. PH2 = 3.92 atm – 2.54 atm = 1.38 atm XH2 = PH2 / Ptot = 1.38 atm / 3.92 atm = 0.352 XO2 = PO2 / Ptot = 2.54 atm / 3.92 atm = 0.648 3 Ex 11 A gas mixture at room temperature contains 4.3 mol of H2 and 8.2 mol of N2. (a) Compute the mole fraction of N2 in the mixture. XN2 = nN2 / ntot = 8.2 mol / (4.3 mol + 8.2 mol) = 0.66 (b) The mixture is then heated, and the N2 starts to react with the H2 to give NH3: N2(g) + 3 H2 (g) → 2 NH3(g) The reaction is stopped when 2.8 mol of NH3 is present. Determine the mole fraction of N2 in this new mixture. mol N2 used = (2.8 mol NH3)(1 mol N2 / 2 mol NH3) = 1.4 mol N2 used mol N2 left over = 8.2 mol – 1.4 mol = 6.8 mol N2 left mol H2 used = (2.8 mol NH3)(3 mol H2 / 2 mol NH3) = 4.2 mol H2 used mol H2 left over = 4.3 mol – 4.2 mol = 0.1 mol H2 left Total mol of gas = 2.8 mol + 6.8 mol + 0.1 mol = 9.7 mol XN2 = nN2 / ntot = 6.8 mol / 9.7 mol = 0.70 Ex 12 Hydrogen gas is produced when zinc reacts with sulfuric acid: Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) If 159 mL of wet H2 is collected over water at 24°C and a barometric pressure of 738 torr, how many grams of Zn have been consumed? (PH2O, 24°C) = 22.38 torr. PH2 = 738 torr – 22.38 torr = 715.62 torr = 0.942 atm nH2 = PV/RT = (0.942 atm)(0.159 L)/R(297K) = 0.006146 mol H2 mass Zn = (0.006146 mol H2)(1 mol Zn/ 1 mol H2)(65.39 g / 1 mol Zn) = 0.402 g Zn 4