Question

Calcium carbonate reacts with hydrochloride acid to produce calcium chloride, CO2 and water. If 50 grams of calcium carbonate was used and ten grams of calcium chloride produced, what was the percentage yield?

Answer 1

The chemical equation for this process is next:

$$\mathrm{CaCO_3} + 2\mathrm{HCl} \rightarrow \mathrm{CaCl_2} + \mathrm{CO_2} + \mathrm{H_2O}$$

As you can see, the molar ratio between calcium carbonate and calcium chloride is 1: 1, it means that one mole of calcium chloride must react with one moles of hydrochloric acid.

Theoretical amount of $\mathrm{CaCO_3}$ is $\mathrm{m/Mw} = 50 / 100 = 0.5$ mol

Real amount of $\mathrm{CaCO_3}$ is equal to calcium chloride amount and it is: $10 / 111 = 0.090$ mol

Yield of reaction is $0.09/0.5 \times 100\% = 18.0\%$