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Nomenclature of Carbonyls

A sample cont...

Question

A sample contains a mixture of $N a H C O_{3}$ and $N a_{2} C O_{3}$ . $H C l$ is added to $15.0$ g of the sample yielding $11.0$ of $N a C l$ . What percent of the sample is $N a_{2} C O_{3}$ ?
Reactions are:
$N a_{2} C O_{3} + 2 H C l \to 2 N a C l + C O_{2} + H_{2} O$
$N a H C O_{3} + H C l \to N a C l + C O_{2} + H_{2} O$
Molecular weight of $N a C l$ , $N a H C O_{3}$ and $N a_{2} C O_{3}$ is $58.5$ , $84$ and $106$ g/mol respectively.

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Solution

Let $x$ g be the weight of of $N a_{2} C O_{3}$ . Then, weight of $N a H C O_{3} = (15 - x)$ g.
Moles of $N a C l$ produced $= \frac{11.0}{58.5} = 0.188$ mol
$N a C l$ is produced by the reaction of $(\frac{x}{106})$ mol of $N a_{2} C O_{3}$ and $\frac{(15 - x)}{84}$ mol of $N a H C O_{3}$ .
Each mol of $N a_{2} C O_{3}$ produces $2$ mol of $N a C l$ .
$∴ \frac{2 x}{106} + \frac{15 - x}{84} = 0.188$
On solving, we get $x = 13.5$ g $N a_{2} C O_{3}$
$N a H C O_{3} = (15 - 1.35) = 13.6$ g
% $N a_{2} C O_{3} = \frac{1.35}{15} \times 100 = 9.0$ % $N a_{2} C O_{3}$
Hence, the percentage of sodium carbonate in the sample is $9.0 %$ .

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