Question

$1.25$ g of sample of limestone on heating gives $0.44$ g carbon dioxide. The percentage purity of $CaC{O}_3$ in limestone is:

• A. $75\%$
• B. $85\%$
• C. $90\%$
• D. $80\%$

Answer 1

Answer: (D)

$80\%$

$CaC{O}_3$ $\underrightarrow{\Delta }CaO+C{O}_2$

Moles of $C{O}_2$ produced =$\frac{0.44}{44}$$=0.01$ moles

Moles of pure $CaCO$${}_3$ required $=0.01$ moles

$=0.01\times 100g$

$=1g$

Therefore $\%$ purity of $CaCO$${}_3$ in limestone =$\frac{1}{1.25}\times 100$

$=80\%$