Question

6.5 g of an impure sample of limestone liberates 2.2 g of CO2 on strong heating. The percentage purity of CaCO3 in the sample is-

1)85.2%
2)76.9%
3)72.5%
4)92.5%

Answer 1

Answer: 76.9%

Explanation:

$CaCO_3\rightarrow CaO+CO_2$

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

$\text{Number of moles of calcium carbonate}=\frac{6.5g}{100g/mol}=0.065 moles$

$\text{Number of moles of carbon dioxide}=\frac{2.2g}{44g/mol}=0.05moles$

According to stoichiometry:

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

Thus 0.065 moles of $CaCO_3$ will give $CO_2$=$\frac{1}{1}\times 0.065=0.065moles$ of $CO_2$

Thus percentage purity= $\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100=\frac{0.05}{0.065}\times 100=76.9\%$

Therefore, the percentage purity of $CaCO_3$ is 76.9%.

Answer 2

Answer:

2)76.9%

Explanation: