Question

The chemical equation shows iron(III) phosphate reacting with sodium sulfate.


2FePO4 + 3Na2SO4 Right arrow. Fe2(SO4)3 + 2Na3PO4


What is the theoretical yield of Fe2(SO4)3 if 20.00 g of FePO4 reacts with an excess of Na2SO4?
26.52 g
53.04 g
150.8 g
399.9 g

Answer 1

Moles FePO4 = 20.00 g / 150.82 g·mol–1 = 0.1326 mol.
Stoichiometry: 2 FePO4 → 1 Fe2(SO4)3, so moles Fe2(SO4)3 = 0.1326/2 = 0.06630 mol.
Molar mass Fe2(SO4)3 ≈ 399.87 g·mol–1, so mass = 0.06630 × 399.87 = 26.52 g.

Answer: 26.52 g.