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Stoichiometric Calculations

A sample of s...

Question

A sample of steel weighing $0.6$ g and containing $S$ as an impurity was burnt in a stream of $O_{2}$ when $S$ was converted to its oxide $S O_{2}$ . $S O_{2}$ was then oxidized to $S {_{4}}^{2 -}$ by using $H_{2} O_{2}$ solution containing $30$ mL of $0.04$ M NaOH. $22.48$ mL of $0.024$ M $H C l$ was required to neutralize the base remaining after oxidation. The $%$ of $S$ in the sample is:

A

1.76 %

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B

1.56 %

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C

1.86 %

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D

1.96 %

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Solution

The correct option is A $1.76 %$
The balanced chemical reaction for oxidation reaction is,
$S O_{2} + H_{2} O_{2} \to H_{2} S O_{4}$
$1 m o l S \equiv 1 m o l S O_{2} \equiv 1 m o l H_{2} S O_{4} \equiv 2 m o l N a O H$

Total number of moles of $N a O H$ $= 0.04 \times \frac{30}{1000} = 0.0012$ mol.

Total number of moles of $H C l$ $= 0.024 \times \frac{22.48}{1000} = 0.00054$ moles.

Total number of moles of $N a O H$ that have reacted with $H_{2} S O_{4} = 0.0012 - 0.00054 = 0.00066$ moles.
Tola number of moles of $H_{2} S O_{4}$ formed $= \frac{0.00066}{2} = 0.00033$ This corresponds to the number of moles of $S$ .
The mass of $S$ is $32 \times 0.00033 = 0.0106$ g.
The percentage of $S$ in the sample is $\frac{0.0106 \times 100}{0.6} = 1.76$ %.

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