Answer

Stoichiometry of Iron and Sulfur Reaction

Reaction Overview

Balanced Equation:

2 Fe (s) + 3 S (s) → Fe2S3 (s)

This is a synthesis reaction forming iron(III) sulfide.

Part 1: Mass of Sulfur to Produce 6.40 Moles of Fe₂S₃

• Molar ratio: 1 mol Fe₂S₃ : 3 mol S
• 6.40 mol Fe₂S₃ × (3 mol S / 1 mol Fe₂S₃) = 19.2 mol S
• Mass = 19.2 mol × 32.07 g/mol = 615.7 g S

Part 2: Mass of Fe₂S₃ from 1.50 g Fe and 1.50 g S

Convert to moles:

• Fe: 1.50 g / 55.85 g/mol = 0.0269 mol
• S: 1.50 g / 32.07 g/mol = 0.0468 mol

Determine the Limiting Reactant:

0.0269 mol Fe × (3 mol S / 2 mol Fe) = 0.0404 mol S required

Available S = 0.0468 mol → S is in excess, Fe is the limiting reactant.

Calculate Fe₂S₃ Formed:

• 0.0269 mol Fe / 2 = 0.01345 mol Fe₂S₃
• Mass = 0.01345 mol × 207.9 g/mol = 2.80 g Fe₂S₃

Part 3: Actual Mass Produced

Since Fe is the limiting reactant, the actual mass of Fe₂S₃ produced is:

Summary Table