Question

PLEASE HELP----
Given the following equation : 2FePo4 + 3Na2SO4 = Fe2(SO4)3 + 2Na3PO4
If I perform this reaction with 25 g of Iron(III) phosphate and an excess of Sodium sulfate, what is my theoretical yeild in grams of Iron (III) sulfate ? If I make 18.5 g of Iron (III) sulfate, what is my percent yeild?

side note i have more questions for points help me

Answer 1

Actual yield of Fe2(So4)3 = 18.5g

2FePo4 + 3Na2SO4 -> Fe2(SO4)3 + 2Na3PO4

Mole of FePO4 = mass of it / its molar mass =
25 g / (55.8 + 31 + 16*4) = 0.166 mol

every 2 mole of FePO4 will form 1 mole of Fe2(SO4)3

Mole of Fe2(SO4)3 produced = 0.166 / 2 = 0.0829 mol

0.0829 * (55.8*2 + 3*(32.1+ 16*4)) = 33.148 g of Fe2(SO4)3

18.5 / 33.148 * 100 = 55.8%