Question

The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 mc010-1.jpg 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

Answer: 26.52g

Solution step-by-step:

1) Chemical equation:

2FePO4 + 3Na2SO4 → Fe2(SO4)3 + 2Na3PO4

2) Theoretical molar ratios:

2 mol FePO4 : 3 mol Na2SO4 : 1 mol Fe2(SO4)3 : 2 mol Na3PO4

3) Convert 20.00 g of FePO4 into number of moles

number of moles = mass in grams / molar mass

molar mass of FePO4 = 150.82 g/mol

numer of moles = 20.00 g / 150.82 g/mol = 0.1326 mol FePO4

4) Proportionality

1 mol Fe2(SO4)3 x
----------------------- = ------------------------
2 mol FePO4 0.1326 mol FePO4

Solve for x:

x = (0.1326 / 2) * 1 mol Fe2(SO4)3 = 0.0663 mol Fe2(SO4)3

5) Convert 0.0663 mol Fe2(SO4)3 into grams

mass in grams = number of moles * molar mass

molar mass Fe2(SO4)3 = 399.88 g/mol

mass = 0.0663 mol * 399.88 g/mol = 26.51 g

Depending on the number of decimal digits used by one or other you might obtain 26.52 instead 26.51, that difference does not count.