Sulfur dioxide reacts with oxygen gas to produce sulfur trioxide .
How many moles of oxygen gas will react with 15.0 L of sulfur dioxide if both gases are at 101.3 kPa and 125 $\omicron $ C?

Question

Quizplus · Accepted Answer

The balanced chemical equation is:2SO2 + O2 -> 2SO3According to the ideal gas law, PV = nRT, the number of moles of a gas is directly proportional to its volume at constant temperature and pressure. Therefore, the mole ratio of the reactants is equal to the volume ratio.Given that the volume of sulfur dioxide is 15.0 L, the volume of oxygen gas required to react with it is:Volume of oxygen gas = (15.0 L) x (1 mole of O2 / 2 moles of SO2) = 7.50 LAt 101.3 kPa and 125°C, the molar volume of a gas is:Molar volume = (RT) / P = [(0.0821 Latm/(molK)) x (398 K)] / (101.3 kPa) = 24.4 L/molTherefore, the number of moles of oxygen gas required is:Number of moles of oxygen gas = (Volume of oxygen gas) / (Molar volume) = (7.50 L) / (24.4 L/mol) = 0.307 molHence, the correct answer is A) 0.230.

Sulfur Dioxide Reacts with Oxygen Gas to Produce Sulfur Trioxide ο\omicron ο

Sulfur Dioxide Reacts with Oxygen Gas to Produce Sulfur Trioxide $\omicron$