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Stoichiometric Calculations

The % of ...

Question

The $%$ of ${(N H_{4})}_{2} C_{2} O_{4}$ is (as nearest integer) :
[Molar mass of ${(N H_{4})}_{2} C_{2} O_{4} = 124.10 g {m o l}^{- 1}$ and atomic mass of $N = 14.0078]$

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Solution

Meq. of ${(N H_{4})}_{2} C_{2} O_{4} =$ Meq. of $N H_{3} =$ Meq. of $H_{2} S O_{4}$ used
$\frac{w}{\frac{124.10}{2}} \times 1000 =$ Meq. of $N H_{3}$
$= (0.05305 \times 2 \times 50 - 11.3 \times 0.1214)$
$= 3.633 g$
$w_{{(N H_{4})}_{2} C_{2} O_{4}} = 0.2273 g$
Percentage of ${(N H_{4})}_{2} C_{2} O_{4} = \frac{0.2273}{0.4775} \times 100$ $= 47.6 %$
So, the nearest integer is $48$ .

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