What is the percent yield for the reaction below when 544.5 g SO2 and 160.0 g O2 produce 382.0 g SO3? 2SO2(g) + O2(g) 2SO3(g) A. 56.1% B. 54.2% C. 23.9% D. 47.7%

Option A.

Explanation:

Balanced equation is:

2SO₂ (g) + O₂(g)  →  2SO₃(g)

First of all, we need to determine the limiting reagent. We convert mass to moles:

544.5 g . 1 mol / 64.06g = 8.5 moles of sulfur dioxide

160 g . 1mol / 32g = 5 moles of oxygen

Ratio is 1:2. 1 mol of oxygen needs 2 moles of sulfur dioxide

5 moles of oxygen may react to (5 . 2) /1 = 10 moles of SO₂

We only have 8.5 moles of SO₂ but we need 10 moles. In conclussion limiting reagent is SO₂.

Ratio is 2:2. 2 moles of SO₂ can prdouce 2 moles of SO₃

Then 8.5 moles of SO₂  must produce 3 moles of SO₃

We convert mass to moles, to determine the theoretical yield (100 % yield reaction) → 8.5 mol . 80.06 g /mol = 680.51 g

Formula for percent yield is: (Produced yield / Theoretical yield) . 100

(382 g / 680.51g) . 100 = 56.1 %

first one is the correct answer