Question

$85\, g \,CaCO _{3}$ (limestone sample), on heating produces exactly the same amount of $CO _{2}$ which converts $30 \,g$ of $MgO$ to $MgCO _{3}$. The percentage purity of limestone sample is

• A. $80 \%$
• B. $82.4\%$
• C. $88.24\% $
• D. $84.8\%$

Answer 1

Answer: (C)

$\underset{40\,g}{MgO}+\underset{44\,g}{CO_{2}} \to MgCO_{3}$

$40 g MgO$ needs $CO _{2}=44 g$

$30 g \,Mg O$ needs $CO _{2}=\frac{44 \times 30}{40}=33 g$

$\underset{100\,g}{CaCO_{3}} \to CaO+\underset{44\,g}{CO_{2}}$

$44 \,g\, CO _{2}$ is obtained from $CaCO _{3}=100 \,g$

$33\, g\, CO _{2}$ is obtained from $CaCO _{3}$

$=\frac{100 \times 33}{44}=75\, g$

Percentage purity of $CaCO _{3}$ sample

$=\frac{75}{85} \times 100=88.24 \%$