Question

A $2.0g$ sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of $C{O}_2$ ceases. The volume of $C{O}_2$, at $750mmHg$ pressure and at $298K$ is measured to be $123.9mL$. A $1.5g$ of the same sample requires $150mL$ of $(M/10)HCI$ for complete neutralisation. Calculate the percentage composition of the components of the mixture.

Answer 1

$C{O}_2$ is evolved due to following reaction :
$2NaHC{O}_3\to N{a}_{2}C{O}_3+H_{2}O+C{O}_2$
Moles of $C{O}_2$ produced $=\frac{pV}{RT}$
$=\frac{750}{760}\times \frac{123.9}{1000}\times \frac{1}{0.082\times 298}$
$=5\times 10^{-3}$
$\Rightarrow$ Moles of $NaHC{O}_3$ in $2g$ sample $=2\times 5\times 10^{-3}=0.01$
$\Rightarrow$ millimol of $NaHC{O}_3$ in $1.5g$ sample
$=\frac{0.01}{2}\times 1.5\times 1000=7.5$
Let the $1.5g$ sample contain $x$ millimol $N{a}_{2}C{O}_3$, then
$2x+7.5=$ millimol of $HCI=15$
$\Rightarrow x=3.75$
$\Rightarrow$ mass of $NaHC{O}_3=\frac{7.5\times 84}{1000}=0.63g$
Mass of $N{a}_{2}C{O}_3=\frac{3.75\times 106}{1000}=0.3975g$
$\Rightarrow \%$ mass of $NaHC{O}_3=\frac{0.63}{1.50}\times 100=42\%$
$\%$ mass of $N{a}_{2}C{O}_2=\frac{0.3975}{1.5}\times 100=26.5\%$