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Stoichiometric Calculations

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A mixture of $H_{2} S O_{4}$ and $H_{2} C_{2} O_{4}$ (oxalic acid) and some inert impurity weighing $3.185 g$ was dissolved in water and the solution made up to $1 L$ , $10 m L$ of this solution required $3 m L$ of $0.1 N N a O H$ for complete neutralization. In another expriment $100 m L$ of the same solution in hot condition required $4 m L$ of $0.02 M K M n O_{4}$ solution for complete reaction. The weight percentage of $H_{2} S O_{4}$ in the mixture was:

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Solution

Let x and y are milli moles $H_{2} S O_{4}$ and $H_{2} C_{2} O_{4}$ in given mixture when both reacted with base
m eq. of acids = m eq. of base;
$(x + y) \times 2 = 3 \times 0.1 \times \frac{1000}{10}$
$\Rightarrow x + y = 15$
In second experiment only $K M n O_{4}$ reacted with $H_{2} C_{2} O_{4}$ in which $M n^{7 +}$ converted into $M n^{2 +}$ and $C_{2} O_{4}^{2 -}$ converted into $C O_{2}$
$∴$
$milli equivalent of H_{2} C_{2} O_{4} = milli equivalent of K M n O_{4}$
$y \times 2 = 4 \times 0.02 \times 5 \times \frac{1000}{100} y = 2 ∴ x = 13, w e i g h t o f H_{2} S O_{4} = 13 \times 10^{- 3} \times 98 = 1.274 g$
weight % of $H_{2} S O_{4}$ in sample $= \frac{1.274}{3.185} \times 100 = 40 %$

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