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%

We are given;

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

The balanced equation for the reaction is given by;

3Zn + Fe₂(SO₄)₃ → 3ZnSO₄ + 2Fe

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

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

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

% 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%

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8

b) nuclear ive learned that in a pedtrodic table