Question

A $10.0g$ a sample of a mixture of calcium chloride and sodium chloride is treated with $N{a}_{2}C{O}_3$ to precipitate 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 $1.62gms$. The $\%$ by mass of $CaC{l}_2$ in the original mixture is :

• A. $15.2\%$
• B. $32.1\%$
• C. $21.8\%$
• D. $11.07\%$

Answer 1

Answer: (B)

$32.1\%$

$CaC{O}_3+heat⟶CaO+C{O}_2$

$X$ $1.68gm$

[molecular mass of $CaC{O}_3=100u$

molecular mass of $CaO=56u$

cross multiply through the equation and solve for $x$ grams of $CaC{O}_3$

$x=\frac{1.62\times 100}{56}$

$x=2.89gm$

$CaC{l}_2+N{a}_{2}C{O}_3⟶CaC{O}_3+2NaCl$

Again cross multiply through the equation and solve for $x$ grams of $CaC{l}_2$

$x=2.89$

$x=\frac{2.89\times 111}{100}$

$x=3.2gm$

$\%CaC{l}_2$ in original mixture $3.2grams$

$CaC{l}_2$ over $10gms$ mixture is $32\%$