Question

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What volume of sulfur dioxide is required to react with 69.2 liters of oxygen gas according to the following reaction? (All gases are at the same temperature and pressure.) sulfur dioxide (g) + oxygen (g) — sulfur trioxide (g) liters sulfur dioxide

Answer 1

Ans 1

The balanced reaction

SO2 + 0.5O2 = SO3

Sulfur dioxide + oxygen = sulfur trioxide

Volume of oxygen gas = 69.2 L

Let pressure = P

Temperature = T

From the ideal gas equation

Moles of oxygen gas n = PV/RT = 69.2P/RT

From the stoichiometry of the reaction

0.5 mol O2 reacts with = 1 mol SO2

69.2P/RT mol O2 reacts with = 69.2P/0.5RT mol SO2

From the ideal gas equation

Volume of SO2 gas = nRT/P

= (69.2P/0.5RT)RT/P

= 138.4 L