Answer to Question #340199 in Chemistry for Andrew

Question

45.2g of Iron metal reacts with excess Oxygen gas to produce 58.1g of

Fe304

a. What is the theoretical yield of Fe304? b. What is the percent yield of this reaction?

Answer 1

Solution:

The molar mass of iron (Fe) is 55.845 g/mol

Therefore,

Moles of Fe = (45.2 g Fe) × (1 mol Fe / 55.845 g Fe) = 0.8094 mol Fe

Balanced chemical equation:

3Fe(s) + 2O2(g) → Fe₃O₄(s)

According to stoichiometry:

3 mol of Fe produce 1 mol of Fe₃O₄

Thus, 0.8094 mol of Fe produce:

(0.8094 mol Fe) × (1 mol Fe₃O₄ / 3 mol Fe) = 0.2698 mol Fe₃O₄

The molar mass of Fe₃O₄ is 231.533 g/mol

Therefore,

Mass of Fe₃O₄ = (0.2698 mol Fe₃O₄) × (231.533 g Fe₃O₄ / 1 mol Fe₃O₄) = 62.4676 g Fe₃O₄

Mass of Fe₃O₄ = 62.4676 g = 62.5 g

The theoretical yield of Fe₃O₄ is 62.5 grams

Percent yield = (Actual yield / Theoretical yield) × 100%

Percent yield = (58.1 g / 62.5 g) × 100% = 92.96%

The percent yield of this reaction is 92.96%

Answers:

(a): The theoretical yield of Fe₃O₄ is 62.5 grams

(b): The percent yield of this reaction is 92.96%