Question

A $2.0\, g$ sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of $CO_2$ ceases. The volume of $CO_2$, at $750 \,mm \, Hg$ pressure and at $298 \,K$ is measured to be $123.9 \,mL$. A $1.5 \,g$ of the same sample requires $150 \,mL$ of $(M/10) HCI$ for complete neutralisation. Calculate the percentage composition of the components of the mixture.

Answer 1

$CO_2$ is evolved due to following reaction :
$2NaHCO_3 \, \rightarrow \, \, Na_2CO_3 + H_2 O + CO_2$
Moles of $CO_2$ 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 $NaHCO_3$ in $2 g$ sample $= 2 \times 5 \times 10^{-3}= 0.01 $
$\Rightarrow $ millimol of $NaHCO_3 $ in $ 1.5\, g $ sample
$ =\frac{0.01}{2}\times 1.5 \times 1000=7.5$
Let the $1.5\, g$ sample contain $x$ millimol $Na_2CO_3$, then
$2x+ 7.5 =$ millimol of $HCI = 15$
$\Rightarrow x=3.75$
$\Rightarrow $ mass of $NaHCO_3 =\frac{7.5 \times 84}{1000}=0.63\, g$
Mass of $Na_2CO_3=\frac{3.75 \times 106}{1000}=0.3975\, g$
$\Rightarrow \%$ mass of $NaHCO_3 =\frac{0.63}{1.50} \times 100 =42 \%$
$\%$ mass of $Na_2CO_2 =\frac{0.3975}{1.5} \times 100 =26.5 \%$