Question

$2 \,g$ sample containing $Na_2CO_3$ and $NaHCO_3$ loses $0.248 \,g$ of mass when heated at $300^{\circ}C$. At this temperature $NaHCO_3$ decomposes to give sodium carbonate in the given sample. What is the $\%$ compostion of the mixture?

• A. $NaHCO_3 \,16.8 \%, Na_2CO_3 \,83.2\%$
• B. $NaHCO_3\, 33.5 \%, Na_2CO_3\, 66.5\%$
• C. $NaHCO_3\, 83.2 \%, Na_2CO_3\, 16.8\%$
• D. $NaHCO_3\, 66.5 \%, Na_2CO_3 \,33.5\%$

Answer 1

Answer: (B)

Weight of the mixture $(NaHCO_3 + Na_2CO_3) = 2 \,g$
$\underset{168\,g}{2NaHCO_3} \xrightarrow{\Delta, 300^{\circ}C} Na_2CO_3 + \underbrace{H_2O \uparrow + CO_2 \uparrow}_{62\,g}$
Let mixture contain $x\, g$ of $NaHCO_3$, then,
Wt. loss from $x\,g$ of $NaHCO_3$
$\Rightarrow \frac{62}{168} \times x = 0.248\,g$
$\Rightarrow x = 0.67\,g$
$ x = \frac{0.248 \times 168}{62} = 0.67\,g$
Then $NaHCO_3$ in sample $ = 0.67\,g$
$\% NaHCO_3 = \frac{0.67}{2} \times 100 = 33.5\%$
and $\% \,Na_2CO_3 = 100 - 33.5$
$ = 66.5\%$