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Test for Collinearity of Vectors

A sample of m...

Question

A sample of mixture of $N a_{2} C O_{3}$ and $N a H C O_{3}$ is subjected to heating till there is no further loss in weight. Assuming that the loss in weight of the sample is $22 %$ of the initial weight of the mixture due to the evolution of $C O_{2}$ , find out the relative percentage of the two components in the mixture.

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Solution

$2 N a H C O_{3} ⟶ N a_{2} C O_{3} + H_{2} O + C O_{2}$
$2$ mol $N a H C O_{3} ⟶ 1$ mol $C O_{2}$
$2 (84 g) ⟶ 44 g$ $C O_{2}$
$44 g m$ of $C O_{2} ⟶ 168 g m$ $N a H C O_{3}$

$N a H C O_{3}$ obtained= $\frac{168}{44} \times 22 = 84 g m$
$N a_{2} C O_{3} = 100 - 84 = 16 g m$
$84$ % of $N a H C O_{3}, 16$ % $N a_{2} C O_{3}$ .

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