Question

30 gram of marble stone on heating produce 11 gram of CO2 . The percentage of CaCo3 in marble is

Answer 1

19% of CaCo3 available in marble after heating

Answer 2

Answer: 83.3%

Explanation:

$CaCO_3\rightarrow CaO+CO_2$

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

$\text{Number of moles of calcium carbonate}=\frac{30g}{100g/mol}=0.3 moles$

$\text{Number of moles of carbon dioxide}=\frac{11g}{44g/mol}=0.25moles$

According to stoichiometry:

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

Thus 0.3 moles of $CaCO_3$ will give $CO_2$=$\frac{1}{1}\times 0.3=0.3moles$ of $CO_2$

Thus percentage purity= $\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100=\frac{0.25}{0.3}\times 100=83.3\%$

Therefore, the percentage purity of $CaCO_3$ is 83.3%.