Answer to Question #340212 in Chemistry for Andrew

Question

45.2g of Iron metal reacts with excess Oxygen gas to produce 58.1g of Fe 3 0 4 a. What is the theoretical yield of F*e_{3}*O_{4} ? 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%