During an experiment, 104 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. if the percent yield of the reaction was 80.15%, what was the actual amount of calcium chloride formed? caco3 + hcl → cacl2 + co2 + h2o 90.1 grams 92.4 grams 109.2 grams 115.3 grams

92.4 grams.

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 = (104 g)/(100.08 g/mol) = 1.039 mol.

Using cross multiplication:

1.0 mole of CaCO₃ produce → 1.0 mole of CaCl₂.

∴ 1.039 mole of CaCO₃ produce → 1.039 mole of CaCl₂.

∴ The amount of CaCl₂ produced = no. of moles x molar mass = (1.039 mol)(110.98 g/mol) = 114.3 g.

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

Percent yield of the reaction = 80.15%, theoretical yield = 115.3 g.

∴ actual yield = [(percent yield of the reaction)(theoretical yield)]/100 = [(80.15%)/(115.3 g)] / 100 = 92.42 g ≅ 92.4 g.

the answer is 1 i believe

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa