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

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.

The answer is the first option just took the test