EXAM PRACTICE QUESTIONS Chapter 12 1. If the volume of a sample of neon is tripled while its Kelvin temperature is quadrupled, what is the affect on the pressure of the gas? If volume is 3V, then the pressure is reduced to 1/3 P (Boyle's Law). If temperature is 4 K, then the new pressure is 4 P (Gay Lusacc's Law). This is a combined gas law problem. 1/3 P u 4 P = 4/3 P, or the net affect on the pressure is an increase by a factor of 1.33. 2. The simplest formula for a compounds is NO2. If 46.0 grams of the gas occupies 11.2 L at STP, the mass of one mole of the compound is ______. The mass of one mole is the same as the MW of the compound. g/molȱ92.0ȱȱ=ȱMWȱL)ȱ(11.2ȱatm)ȱ(1.00K)ȱ)(273KmolatmLȱg)(0.0821ȱ(46.0ȱ=ȱMWȱȱȱȱPVgRTȱ=ȱMWȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱMWgRTȱ=ȱPV? 3. The volume of a sample of gas is 405 mL at 10.0 atm and 467 K. What volume will it occupy at 4.29 atm and the same temperature? This is a Boyle's Law problem, P1V1 = P2V2. mLȱ944ȱ=ȱVȱatmȱ4.29atm)ȱ(10.0ȱmL)ȱ(405ȱ=ȱPVPȱ=ȱV22112 4. If 5.50 L of argon gas at 273 qC is heated at constant pressure to a temperature of 546 qC, what would be its new volume? This is a Charles' Law problem. Lȱ8.25ȱ=ȱK)ȱ(546K)ȱ)(Lȱ(5.50ȱ=ȱVTTVȱ=ȱVȱȱȱȱȱȱȱȱȱȱȱȱTVȱ=ȱTV212122211819?5. A sample of helium occupies 1.40 L at standard conditions. What pressure will it exert in a 5.00 L vessel at 100. qC? V1P1T1 = V2P2T2 ? P2 = V1P1T2T1V2P2 = (1.40 L)(760 torr)(373 K)(273 K)(5.00 L) = 291 torr 6. Given the mass of one mole of gaseous NO is 30.0 grams/mol, calculate the density of NO in grams/liter at STP. g/Lȱ1.34ȱ=ȱVgȱK)ȱ)(273KmolatmLȱ(0.0821)molgȱ(30.0ȱatmȱ1.00ȱ=ȱRTP(MW)ȱ=ȱVgLgȱ=ȱVgȱ=ȱȱȱȱȱȱȱȱȱDMWgRTȱ=ȱPV(density) 7. A dentist has a 5.00 L cylinder of "laughing gas", N2O, at 20.0 atm and 25 qC. What mass of gas is contained in the cylinder? gȱ180ȱ=ȱK)ȱ)(298KmolatmLȱ(0.0821)molgȱ(44.0ȱL)ȱ(5.00ȱatmȱ20.0ȱ=ȱgRTPV(MW)ȱ=ȱgȱȱȱȱȱȱȱȱȱȱȱȱȱȱMWgRTȱ=ȱPV? 8. A 10.0 L flask contains 0.400 mol of H2, 0.300 mole of He and 0.500 mole of Ne at 35.0 qC. What is the partial pressure of the Ne? 1. Use Dalton's Law of Partial Pressures to find the total moles of gas. 0.400 mol H2 + 0.300 mol He + 0.500 mol Ne = 1.20 mol gas 2. We use the ideal gas equation to find the total pressure of the gas mixture. atmȱmolȱ1.26ȱ=ȱ)ȱmolȱ1.20Neȱmolȱ0.500(ȱatmȱ3.03ȱ=ȱPatmȱ3.03ȱ=ȱL)ȱ(10.0ȱK)ȱ(308ȱ)KmolatmLȱ(0.0821ȱmol)ȱ(1.20ȱ=ȱPVnRT)ȱ=ȱPȱȱȱȱȱȱȱȱȱȱȱȱȱȱnRTȱ=ȱPVNe?9. A mixture of gases containing 26.0 g SO2, 4.00 g H2, and 20.2 g Ne at 23 qC is in a 5.00 L container. What is the partial pressure of H2? Use Dalton's Law of Partial Pressures to find the total moles of gas. gasȱofȱmolȱtotalȱ.ȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱȱHeȱmolȱ1.00ȱ=ȱHeȱgȱ20.2Heȱmolȱ1ȱȱHeȱgȱ20.2Hȱmolȱ2.00ȱ=ȱHȱgȱ2.00Hȱmolȱ1ȱȱHȱgȱ4.00SOȱmolȱ0.406ȱ=ȱSOȱgȱ64.1SOȱmolȱ1ȱȱSOȱgȱ26.022222222413uuu We use the ideal gas equation to find the total pressure of the gas mixture. atmȱ16.5ȱ=ȱL)ȱ(5.00ȱK)ȱ(296ȱ)KmolatmLȱ(0.0821ȱmol)ȱ(3.41ȱ=ȱVnRTȱ=ȱP We determine the partial pressure of H2 by multiplying the XH2 (mol fraction) by the total gas pressure. atmȱ9.70ȱ=ȱatm)ȱ(16.5ȱ)gasȱmolȱ3.41Hȱmolȱ2.00(ȱ=ȱ)(PXȱ=ȱP2totalHH22 10. A 500.-mL sample of O2 was collected over water at 24 qC under a barometric pressure of 738 torr. What mass of dry oxygen was collected? The vapor pressure of water at 24 qC is 22 torr. 1. The total pressure is due to oxygen plus water. We first determine the partial pressure of O2. PO2 = Ptotal - PH2O = 738 - 22 torr = 716 torr 2. We use the ideal gas equation to determine the mass of oxygen. 2Oȱgȱ0.619ȱ=ȱK))(297ȱKmolatmLȱ(0.0821)molgȱ(32.0ȱL)ȱ(0.500ȱ)torrȱ760atmȱ1ȱȱtorrȱ(716ȱ=ȱgRTPV(MW)ȱ=ȱgȱȱȱȱȱȱȱȱȱȱȱȱȱȱMWgRTȱ=ȱPVu? 11. What volume of hydrogen (at STP) is produced by dissolving 0.500 mol of Al in H2SO4? 2 Al(s) + 3 H2SO4(aq) o Al2(SO4)3(aq) + 3 H2(g) Lȱ16.8ȱȱHȱmolȱ1HL.ȱȱAlȱmolȱ2Hȱmolȱ3ȱȱAlȱmolȱ0.500222 uu42212. What is the order of increasing rate of effusion for the following gases: Ar, CO2, He and N2. Rates of effusion are indirectly related to molecular weight. In other words, the higher the molecular weight of a molecule, the slower it moves and effuses. MW1ȱȱeffusionȱofȱrate v CO2 < Ar < N2 < He slowest fastest 13. What is the molecular weight of a (hypothetical) gas that diffuses 1.414 times faster than N2? g/molȱ14.0ȱ=ȱ2.00g/molȱ28.0ȱ=ȱUnkȱMWUnkȱMWg/molȱ28.0ȱ=ȱ.UnkȱMWg/molȱ28.0ȱ=ȱ1.414UnkȱMWg/molȱ28.0ȱ=ȱ1.414ȱUnkȱMWNȱȱMWȱ=ȱ1ȱȱȱNȱRate1.414ȱȱȱUnkȱRate220022¸¸¹·¨¨©§ 14. 8.00 g of H2 gas is reacted with 0.400 mol of O2 gas at STP according to the following equation. 2 H2(g) + O2(g) o 2 H2O(g)a) Calculate the volume, in L, of H2O produced in the reaction. Determine the limiting reagent. OH ȱLȱ89.6ȱȱOH ȱmolȱ1OHL22.4ȱȱȱHȱmolȱ2OH ȱmolȱ2ȱȱȱHȱgȱ2.00Hȱmolȱ1ȱȱ gȱ8.002222222 uuu OH ȱLȱ17.9ȱȱOH ȱmolȱ1OHL22.4ȱȱȱOȱmolȱ1OH ȱmolȱ2ȱȱȱOȱmolȱ0.400222222 uu O2 is limited because it produces the least water, 17.9 L, which is the amount of water produced in the reaction.b) Calculate the volume, in L, of the excess reactant that remains. First, calculate the amount of H2 available. availableȱHȱȱmolȱ4.00ȱȱȱHȱgȱ2.00Hȱmolȱ1ȱȱHȱgȱ8.002222 u Next calculate the H2 used based (use the limiting reagent oxygen to determine) reactionȱinȱusedȱHȱmolȱ0.800ȱȱȱOȱmolȱ1Hȱmolȱ2ȱȱȱOȱmolȱ0.4002222 u 4.00 mol H2 - 0.800 mol H2 = 3.20 mol H2 remain excessȱinȱemainrȱHȱLȱ71.9ȱȱȱȱHȱmolȱ1HȱLȱ22.4ȱȱȱHȱmolȱ3.202222