An experiment reacts 20.4 g of zinc metal with a solution containing an excess of iron (iii) sulfate. after the reaction, 10.8 grams of iron metal are recovered. what is the percent yield of the experiment?

The answer to your question is 92.7%

Explanation:

Balanced Chemical reaction

                             3 Zn  + Fe₂(SO₄)₃   ⇒   2Fe   +   3ZnSO₄

Molecular weight

Zinc = 65.4 x 3 = 196.2g

Iron (III) = 56 x 2 = 112 g

Proportions  

                           196.2 g of Zinc ------------------ 112 g of Iron

                            20.4 g of Zinc  -----------------   x

                            x = (20.4 x 112) / 196.2

                            x = 2284.8/196.2

                            x = 11.65 g of Iron

% yield =

% yield = 0.927 x 100

% yield = 92.7

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the answer to this question is 3.