** will give you 25 points plus brainliest ** during an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. if the percent yield of the reaction was 82.15%, what was the actual amount of calcium chloride formed? caco3 + hcl → cacl2 + co2 + h2o a. 105.3 grams b. 101.1 grams c. 95.6 grams d. 86.5 grams

D. 86.5 grams

Explanation:

∵ The percentage yield = (actual yield/theoretical yield)*100.

From the balanced reaction:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O,

It is clear that 1 mol of CaCO₃ reacts with 2 mol of HCl to produce 1 mol of CaCl₂, 1 mol of CO₂, and 1 mol H₂O.

n = mass/molar mass = (95.0 g)/(100.0869 g/mol) = 0.95 mol.

Using cross multiplication:

1 mol of CaCO₃ produce  → 1 mol of CaCl₂, from stichiometry.

∴ 0.95 mol of CaCO₃ produce → 0.95 mol of CaCl₂.

∴ The mass of CaCl₂ (theoretical yield) = (no. of moles) * (molar mass) = (0.95 mol)*(110.98  g/mol) = 105.34 g.

∵ The percentage yield = (actual yield/theoretical yield)*100.

The percentage yield = 82.15%, theoretical yield = 105.34 g.

∴ The actual yield of CaCl₂ = (The percentage yield)(theoretical yield)/100 = (82.15%)(105.34 g)/100 = 86.53 g ≅ 86.5 g.

11 players on a field that has a maximum of 11 players

bcde is the incorrect answer it is a