Gas Stoichiometry

You know the drill… 1.Write the balanced chemical equation 2.State your givens Make sure your units are consistent (i.e. convert to Kelvin, kPa, etc.) 3.Use mole ratios to solve

Example 1: Volume-to-Volume A catalytic converter in the exhaust system of a car uses oxygen (from the air) and a catalyst to convert poisonous carbon monoxide to carbon dioxide. If temperature and pressure remain constant, what volume of oxygen is required to react with 65.0 L of carbon monoxide produced during a road trip? 2 CO (g) + O 2(g)  2 CO 2(g) 65.0 LV = 65.0 L X 1 mol O2 2 mol CO = 32.5 L Law of Combining Volumes: At a constant T and P, volumes of gaseous reactants and products react in whole-number ratios.  The volume of oxygen required is 32.5 L

Example 2: Mass-to-Volume What volume of carbon dioxide is produced when 6.40 g of methane gas, CH 4, reacts with excess oxygen? All gases are at 35.0˚C and kPa. CH 4(g) + 2 O 2(g)  CO 2(g) +2 H 2 O (g) m = 6.40 g MM = g/mol n = 6.40 g X 1 mol g n = mol n = mol CH4 X 1 mol CO2 1 mol CH4 n = mol CO2 PV = nRT T = K R = kPa·L/mol·K PV = nRT V = nRT P V = ( mol)(8.314 kPa·L·mol -1 ·K -1 )( K) kPa V = 10.2 L

Example 3: Volume-to-Mass What mass of sodium azide, NaN 3(s), is required to produce the 67.0 L of nitrogen gas that is needed to fill a car’s airbag? Assume a temperature of 32˚C and a pressure pf 105 kPa 2 NaN 3(s)  2 Na (s) +3 N 2(g) n = mol N2 X 2 mol NaN3 3 mol N2 n = mol NaN3 MM = g/mol m = mol X g 1 mol m = g m = 1.2 X 10 2 g V = 67.0 L PV = nRT n = PV RT n = (105 kPa)(67.0 L) (8.314 kPa·L·mol -1 ·K -1 )( K) n = mol PV = nRT T = K R = kPa·L/mol·K