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Gas Stoichiometry.

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Presentation on theme: "Gas Stoichiometry."— Presentation transcript:

1 Gas Stoichiometry

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

3 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? Law of Combining Volumes: At a constant T and P, volumes of gaseous reactants and products react in whole-number ratios. 2 CO(g) + O2(g)  2 CO2(g) 65.0 L V = 65.0 L X 1 molO2 2 molCO = 32.5 L  The volume of oxygen required is 32.5 L

4 Example 2: Mass-to-Volume
PV = nRT What volume of carbon dioxide is produced when 6.40 g of methane gas, CH4, reacts with excess oxygen? All gases are at 35.0˚C and kPa. T = K R = kPa·L/mol·K CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) m = 6.40 g MM = g/mol n = 6.40 g X 1 mol 16.05 g n = mol n = molCH4 X 1 molCO2 1 molCH4 n = molCO2 PV = nRT V = nRT P V = ( mol)(8.314 kPa·L·mol-1·K-1)( K) 100.0 kPa V = 10.2 L

5 Example 3: Volume-to-Mass
PV = nRT What mass of sodium azide, NaN3(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 T = K R = kPa·L/mol·K 2 NaN3(s)  2 Na(s) + 3 N2(g) n = molN2 X 2 molNaN3 3 molN2 n = molNaN3 MM = g/mol m = mol X g 1 mol m = g m = 1.2 X 102 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


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