Question

A $10.0g$ sample of a mixture of calcium chloride and sodium chloride is treated with $N{a}_{2}C{O}_3$ to precipiate the calcium as calcium carbonate. This $CaC{O}_3$ is heated to convert all the calcium to $CaO$ and the final mass of $CaO$ is $1.62g$. The $\%$ by mass of $CaC{l}_2$ in the original mixture is

• A. $15.2\%$
• B. $32.1\%$
• C. $21.8\%$
• D. $11.7\%$

Answer 1

The correct option is B $32.1\%$

Let the mass of $CaC{l}_2$ in sample $=$ $xg$ $CaC{l}_2+N{a}_{2}C{O}_3\to CaC{O}_3+2NaCl$
$111$ g $100$ g
$\therefore 111gCaC{l}_2\text{produces}100gCaC{O}_3$
Hence $xgCaC{l}_2\text{produces}\frac{100x}{111}gCaC{O}_3$
$CaC{O}_3\underrightarrow{\Delta }CaO+C{O}_2$
$100g$ $56g$
$\therefore 100gCaC{O}_3\text{produces}56gCaO$
$\left(\frac{100x}{111}\right)gCaC{O}_3\text{produces}\frac{56}{100}\times \frac{100}{111}xgCaO=\frac{56x}{111}gCaO$
Since mass of $CaO$ finally produced $=1.62g\Rightarrow \frac{56}{111}x=1.62g$
$\Rightarrow x=3.21g$
$\%$ of $CaC{l}_2$ in sample $=\frac{3.21}{10}\times 100=32.1\%$