Sulfur dioxide reacts with oxygen gas to produce sulfur trioxide .
What volume 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

We can use the equation: 
SO2 + O2 → SO3

According to the balanced equation, the stoichiometric ratio of SO2 to O2 is 2:1. This means that for every 2 moles of SO2, we need 1 mole of O2. 
We can use the ideal gas law: 
PV = nRT

to find the number of moles of each gas. Since both gases have the same pressure, temperature, and volume, we can use the same values of P, V, and T for both gases.

nSO2 = (PV)/(RT) = [(101.3 kPa)(15.0 L)]/[(0.08206 L·atm/mol·K)(398 K)] = 0.573 mol

nO2 = (1/2)nSO2 = 0.2865 mol

Now we can use the ideal gas law again to find the volume of O2:

V = (nRT)/P = [(0.2865 mol)(0.08206 L·atm/mol·K)(398 K)]/(101.3 kPa) = 7.50 L

Therefore, the answer is C) 7.50 L.

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

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