Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

Answer 1

Answer: a) 18.3 grams

b) $CaCO_3$ is the excess reagent and 16.5g of $CaCO_3$ will remain after the reaction is complete.

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of calcium carbonate}=\frac{26g}{100g/mol}=0.26moles$

$\text{Number of moles of hydrochloric acid}=\frac{12g}{36.5g/mol}=0.33moles$

According to stoichiometry:

2 mole of $HCl$ react with 1 mole of $CaCO_3$

0.33 moles of $HCl$ will react with=$\frac{1}{2}\times 0.33=0.165moles$ of $CaCO_3$

Thus $HCl$ is the limiting reagent as it limits the formation of product and $CaCO_3$ is the excess reagent.

2 moles of $HCl$ produce = 1 mole of $CaCl_2$

0.33 moles of $HCl$ produce=$\frac{1}{2}\times 0.33=0.165moles$ of $CaCl_2$

Mass of $CaCl_2=moles\times {\text{Molar Mass}}=0.165\times 111=18.3g$

As 0.165 moles of $CaCO_3$ are used and (0.33-0.165)=0.165 moles of $CaCO_3$ are left unused.

Mass of $CaCO_3$ left unreacted =$moles\times {\text {Molar mass}}=0.165\times 100=16.5g$

Thus 18.3 g of $CaCl_2$ are produced. $CaCO_3$ is the excess reagent and 16.5g of $CaCO_3$ will remain after the reaction is complete.