an experiment reacts 21 grams of zinc metal with a solution of iron 3 sulphate. after the reaction, 10.8 grams of iron metal are recovered. what is the percent yield of the experiment?

90.38%

Explanation:

We are given;

Mass of Zinc metal is 21 g Mass of Iron metal produced is 10.8 g

We are required to calculate the percent yield of the experiment;

Step 1: Write the balanced equation for the reaction.

The balanced equation for the reaction is given by;

3Zn + Fe₂ (SO₄) ₃ → 3ZnSO₄ + 2Fe

Step 2: Moles of Zinc metal used

We know that, moles = Mass : Molar mass

Molar mass of Zn is 65.38 g/mol

Therefore;

Moles of Zn = 21 g : 65.38 g/mol

= 0.321 moles

Step 2: Moles of Fe produced From the equation;

3 moles of Zinc metal reacts to yield 2 moles of iron metal

Therefore; Moles of Iron = Moles of Zinc * 2/3

= 0.321 moles * 2/3

= 0.214 mol

Step 3: Theoretical yield of the reaction

Theoretical mass of Iron metal produced

Mass = Number of moles * Molar mass

Molar mass of iron is 55.845 g/mol

Thus, mass of Iron = 0.214 mol * 55.845 g/mol

= 11.95 g

Step 4: Percent yield of the experiment

% yield = (Actual yield : theoretical yield) * 100

Therefore;

% yield = (10.8 g : 11.95 g) * 100

= 90.38%

Thus, the % yield of the experiment is 90.38%