Question

$50g$ of an impure calcium carbonate sample decomposes on heating to give carbon dioxide and $22.4g$ calcium oxide. The percentage purity of calcium carbonate in the sample is:

• A. $60\%$
• B. $80\%$
• C. $90\%$
• D. $70\%$

Answer 1

Answer: (B)

$80\%$

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

Let pure sample of $CaC{O}_3=x$ $grams$

Mass of $CaO$ produced after decomposition $=22.4$ $g$

Molar mass of $CaC{O}_3=100$ $g/mol$

and, Molar mass of $CaO=56$ $g$

If $100\%$ is pure, then

$100$ $g$ $CaC{O}_3⟶56$ $g$ of $CaO$

Also,$y$ $g$ of $CaC{O}_3⟶22.4$ $g$ of $CaO$

Dividing these two,

$\frac{100}{y}=\frac{56}{22.4}$

$y=40$ $grams$

$\therefore$ Percentage of purity $=\frac{Massofpuresample}{Totalmassofimpure}\times 100=\frac{40}{50}\times 100=80\%$