Question

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?

Answer 1

Answer:

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 = $\frac{10.8}{11.65} x 100$

% yield = 0.927 x 100

% yield = 92.7

Answer 2

Answer:

93.1%

Explanation:

As with all stoichometry problems, you must start by writing a balanced equation.

3 Zn (s) + Fe2O3 (s) → 2 Fe (s) + 3 ZnO (s)

Since you are given that there is 20.4g of Zinc metal as a reactant, that is your given. You will use this number to calculate your theoretical yield by using the mole ratio between Zinc and Iron.

20.4 g Zn • 1 mol Zn/65.38 g Zn • 2 mol Fe/3 mol Zn • 55.85g Fe/1 mol Fe = 11.6g Fe

Therefore, your theoretical yield is 11.6g.

Now, you divide your actual yield (10.8g) by your theoretical yield (11.6g) and multiply by 100 to get a percentage.

10.8g/11.6g • 100 = 93.1%

Therefore, your percent yield is 93.1%.