Question

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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

Answer 1

Answer:

D. 86.5 grams

Explanation:

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

• We need to calculate the theoretical yield:

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.

• We need to calculate the no. of moles of 95.0 g CaCO₃:

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.

• So, the right choice is: D. 86.5 grams.

Answer 2

Answer:

The correct answer is option D.

Explanation:

$CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$

Moles of calcium carbonate = $\frac{95 g}{100.08 g/mol}=0.9492 mol$

According to reaction,1 mol of calcium carbonate gives 1 mol of calcium chloride.

Then 0.9492 mol of calcium carbonate will give :

$\frac{1}{1}\times 0.9492 mol=0.9492 mol$ of calcium chloride.

Theoretical yield of the calcium chloride = $110.98 g/mol\times 0.9492 mol=105.3422 g$

Experimental yield of calcium chloride = 95 g

Percentage yield of a calcium chloride = 82.15 %

$\frac{\text{experimental yield}}{\text{Theoretical yield}}\times 100=82.15\%$

$\frac{\text{experimental yield}}{105.3422 g}\times 100=82.15\%$

Experimental yield = 86.53 g

The closest option to our answer is option D. Hence, option D is correct answer.