Quiz on Gases-Problems

Practice Quiz For Chapter 13; Gases Data: 1.0 atm = 760 torr 1.0 atm = 101,325 Pa Molar masses: O 15.9 N 14.01 Ar 39.95 C 12.01 H 1.01 R= 0.08206 L atm / K mol n = mass /Molar mass STP = 0 C and 1.00 atm Temp. K = Temp. C + 273 1.What is the volume of CO2 (g) measured at STP if you have 2.00g of CH4? CH4(g) + 2O2(g)  CO2(g) + 2H2O (g) 1.What is the volume of CO2 (g) measured at STP if you have 2.00g of CH4? CH4(g) + O2(g)  CO2(g) + 2H2O (g) Balanced = CH4(g) + 2O2(g)  CO2(g) + 2H2O (g) T = 273 K P = 1.00 atm mass (CH4) = 2.00g PV = nRT V =? mass(g) 2.00g n(CH 4 )  1  1  0.125mol molarmass(gmol ) 16.039gmol n(CH 4 )  n(CO2 )  2.80L From equation nRT (0.125mol)(0.08206atmL/ molK)(273K) V  P (1.00atm) 2.A 10.0L tank contains 3.20g of CO2 (g) and 5.00g of O2 (g) at 25C What is the total pressure inside of the tank? V = 10.0L P = ? atm mass (CO2) = 3.20g PV = nRT T = 25 C = 298 K mass (O2) = 5.00g P= n(CO 2) nRT V mass(g) 3.20g  1  1  0.0730mol molarmass(gmol ) 43.81gmol mass(g) 5.00g n(O2 )  1  1  0.157mol molarmass(gmol ) 31.8gmol n(total)  n(CO2 )  n(O2 )  0.230mol ntotal RT (0.230mol)(0.08206atmL/ molK)(298K) P  V (10.0L)  0.562atm 3. A gas occupies 18.0 L at 860 torr and 38 C, What temperature is necessary to increase the volume to 30.0L if the pressure stays the same? V1 = 18.0L P = 860 torr = 860torr/760torr = 1.132 atm V2 = 30.0L T1 = 38C = 311 K T2 = ? Pressure stays constant, assume the number of moles is  P  constant V  nRT V  RT V R T V1 V2 R T1 T2  PV  nRT  V  RT V R T V1 V2 R T1 T2 V1 V2 18.0L 30.0L    T1 T2 311K T2 (30.0L)(311K) T2  18.0L  518K 4. An oxygen sample has a volume of 4.50 L at 25 C and 790.0 torr. How many moles of oxygen molecules does it contain? V = 4.50 L T = 25 C = 298 K P = 790 torr = 1.039 atm PV = nRT n=PV RT n= (1.039atm) (4.50L) (0.08206L atm K-1 mol -1 )(298 K) 4.6755 atm L 24.45 L atm K-1 mol -1 K n= n= 0.191 moles 5. Convert the following: a. 898.5 mm Hg to atm b. 0.408 atm to torr c. 68,471 Pa to mm Hg 1.0 atm = 760 torr = 760 mm Hg 1.0 atm = 101,325 Pa 1.0 atm = 14.69 psi d. 50.9 psi to atm a. 898.5mmHg  1.00atm  1.18atm 760mmHg 760torr b. 0.408atm  1.00atm  310.08torr 760mmHg 68,471Pa   514mmHg 101,325Pa c. d. 1.00atm 50.9psi   3.46atm 14.69 psi 6. A gas has a pressure of 2,500 psi, how many torr is this? Hint 1.000 atm = 14.69 psi 760torr 2500psi   129,339.68torr 14.69 psi or 1.00atm 2500psi   170.18atm 14.69 psi 760torr 170.18atm   129,339.68torr 1.00atm 7. A sample of nitrogen gas at 2.4atm has a volume of 50.3 L. If the pressure is decreased to 1.9 atm, will the new volume be greater or smaller? What is the new volume? P V1  P2 V2 1 (2.4atm)(50.3L)  (1.9atm)(V2 ) P1V1 V2  P2 (2.4atm)(50.3L) V2  (1.9atm) V2  63.5L This is using Boyle’s Law 8. A container with a moving piston contains 0.89 L of methane gas at 100.5˚C. If the temperature of the gas rises by 11.3 ˚C, what is the new volume of the gas? PV  nRT   V2 = 0.99L V  constant T V1 V2  T1 T2 0.89L V2  100.5 111.8 V2  (0.89)(111.8)/ 100.5 (0.89)(111.8) V2  (100.5) 9. Carbon dioxide is produced during the combustion of liquid propane fuel. C3H8 + O2 3CO2 + 4H2O Balanced: C3H8 +5O2 3CO2 + 4H2O If 5.0kg of propane are burned at 1.000 atm and 400 ˚ C, what volume of CO2 gas is produced? Mass propane = 5000g n(propane) = 5000g/44.11 = 113.35 moles n(CO2) = 3 (113.35) =340.06 moles PV =nRT nRT V P (340.05)(0.08206)(673K) V (1.00atm) V =18,780.22L