During an experiment, 102 grams of calcium carbonate react with an excess amount of hydrochloric acid. if the percent yield of the reaction was 85.15%, what was the amount of calcium chloride formed?

96.31 g.

Explanation:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O,

1.0 mole of CaCO₃ reacts with 2.0 moles of HCl to produce 1.0 mole of CaCl₂, 1.0 mole of CO₂, and 1.0 mole of H₂O.

no. of moles of CaCO₃ = mass/molar mass = (102 g)/(100.08 g/mol) = 1.019 mol.

Using cross multiplication:

1.0 mole of CaCO₃ produce → 1.0 mole of CaCl₂.

∴ 1.019 mole of CaCO₃ produce → 1.019 mole of CaCl₂.

∴ The amount of CaCl₂ produced = no. of moles x molar mass = (1.019 mol)(110.98 g/mol) = 113.1 g.

∵ percent yield of the reaction = [(actual yield)/(theoretical yield)] x 100.

Percent yield of the reaction = 85.15%, theoretical yield = 113.1 g.

∴ actual yield = [(percent yield of the reaction)(theoretical yield)]/100 = [(85.15%)(113.1 g)] / 100 = 96.31 g.

it is b that is the answer.