What volume of sulfur dioxide gas at 45 °C and 723 mmHg will react completely with 0.870 L of oxygen gas at constant temperature and pressure? 2 SO₂(g)+ O₂(g)$\rarr$2SO₃(g) A)0.0317 L B)0.0634 L C)0.870 L D)1.74 L E)3.48 L

Question

Quizplus · Accepted Answer

We can use the equation and stoichiometry to find the amount of sulfur dioxide gas needed to react completely with 0.870 L of oxygen gas. The balanced equation tells us that 2 moles of sulfur dioxide react with 1 mole of oxygen gas to produce 2 moles of sulfur trioxide gas. Therefore, we can set up a proportion:

2 mol SO2 / 1 mol O2 = x L SO2 / 0.870 L O2

Solving for x, we get:

x = 0.435 mol SO2

Now we can use the ideal gas law to find the volume of sulfur dioxide gas:

PV = nRT

V = nRT/P

V = (0.435 mol)(0.0821 L atm/mol K)(318 K) / 723 mmHg

V = 0.0317 L

Therefore, the answer is A) 0.0317 L of sulfur dioxide gas.

What Volume of Sulfur Dioxide Gas at 45 °C and 723