Question

50 gram sample of CaCO3 on heating gives 17.6 g of CO2 calculate percentage purity ​

Answer 1

Answer:  The percentage purity of $CaCO_3$ is 80%.

Explanation:

$CaCO_3\rightarrow CaO+CO_2$

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

$\text{Number of moles of calcium carbonate}=\frac{50g}{100g/mol}=0.5 moles$

$\text{Number of moles of carbon dioxide}=\frac{17.6g}{44g/mol}=0.4moles$

According to stoichiometry:

1 mole of $CaCO_3$ gives 1 mole of $CO_2$

Thus 0.5 moles of $CaCO_3$ will give $CO_2$=$\frac{1}{1}\times 0.5=0.5moles$ of $CO_2$

Thus percentage yield = $\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100=\frac{0.4}{0.5}\times 100=80\%$

Therefore, the percentage purity of $CaCO_3$ is 80%.