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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?

Respuesta :

nkirotedoreen46

Answer:

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%

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