Question

$21.2\text{g}$ sample of impure $N{a}_{2}C{O}_3$ is dissolved and reacted with a solution of $CaC{l}_2$, the weight of $CaC{O}_3$ formed was $10.0\text{g}$. Which of the following statements is/are correct?

• A. The percentage purity of $N{a}_{2}C{O}_3$ is $50\%$
• B. The percentage purity of $N{a}_{2}C{O}_3$ is $60\%$
• C. The number of moles of $N{a}_{2}C{O}_3=CaC{O}_3=0.1\text{mol}$
• D. The number of moles of $NaCl$ formed is $0.1\text{mol}$

Answer 1

Answer: (A), (C)

The correct options are
A The percentage purity of $N{a}_{2}C{O}_3$ is $50\%$
C The number of moles of $N{a}_{2}C{O}_3=CaC{O}_3=0.1\text{mol}$

Given: Mass of $N{a}_{2}C{O}_3=21.2\text{g}$
Mass of $CaC{O}_3=10\text{g}$
Molar mass of $N{a}_{2}C{O}_3=(23\times 2)+12+(16\times 3)=106\text{g}$
Molar mass of $CaC{O}_3=40+12+(16\times 3)=100\text{g}$
Let, mass of impurity $=x\text{g}$
Balanced chemical reaction is
$N{a}_{2}C{O}_3+CaC{l}_2\to CaC{O}_3+2NaCl$
From stoichiometry:
Moles of $N{a}_{2}C{O}_3=$ Moles of $CaC{O}_3$
Since, moles = mass/Molar mass
$\frac{21.2-x}{106}=\frac{10}{100}$
$212-10x=106$
$x=10.6\text{g}$
$\%$ purity of $N{a}_{2}C{O}_3=\frac{\text{Mass of pure}N{a}_{2}C{O}_3}{\text{Mass of impure}N{a}_{2}C{O}_3}\times 100$
$=\frac{10.6}{21.2}\times 100=50\%$
Moles of $N{a}_{2}C{O}_3=\frac{21.2-x}{106}=\frac{21.2-10.6}{106}=0.1$
Moles of $CaC{O}_3=\frac{10}{100}=0.1$
From stoichiometry,
$\frac{\text{moles of}N{a}_{2}C{O}_3}{1}=\frac{\text{moles of}NaCl}{2}$
moles of $NaCl=0.1\times 2=0.2$
So, options a and c are correct.