Question

$10$ g of a sample of a mixture of $CaC{l}_2$ and $NaCl$ is treated to precipitate all the calcium as $CaC{O}_3$. This $CaC{O}_3$ is heated to convert all the $Ca$ to $CaO$ and the final mass of $CaO$ is $1.62$ g. The percent by mass of $CaC{l}_2$ in the original mixture is :

• A. $32.1\%$
• B. $16.2\%$
• C. $1.8\%$
• D. $11.0\%$

Answer 1

The correct option is A $32.1\%$

Weight of ($CaC{l}_2+NaCl)=10$ g

Let the weight of $CaC{l}_2=x$ g

Molecular mass of $CaC{l}_2=40+(35.5\times 2)=40+71=111$

Moles of $CaC{l}_2=\frac{x}{111}moles$

One mole of $CaC{l}_2$ produce one mole of $CaO$.Hence,

$\underset{\underset{\frac{x}{111}mol}{1mol}}{CaC{l}_2}\to \underset{\underset{\frac{x}{111}mol}{1mol}}{CaC{O}_3}\to \underset{\underset{\frac{x}{111}mol}{1mol}}{CaO}$

The molecular mass of $CaO$=56g

Moles of $CaO=\frac{1.62}{56}=0.028moles$

$\therefore \frac{x}{111}=0.028$

$\therefore x=3.21$ g

% of $CaC{l}_2=\frac{3.21}{10}\times 100=$ $32.1\%$.

Hence, option $A$ is correct.