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Percentage Composition

The volume (i...

Question

The volume (in ml) of $6 M$ $N a O H$ required to react with the $S O_{2}$ which is produced from $10^{3}$ kg of coal containing $0.05 %$ by mass of $F e S_{2}$ (iron pyrites) is :
$2 F e S_{2} + 5.5 O_{2} \to F e_{2} O_{3} + 4 S O_{2}$
$S O_{2} + 2 N a O H \to N a_{2} S O_{3} + H_{2} O$

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Solution

The mass of $F e S_{2} = 1000000 g \times \frac{0.05}{100} = 500$ g.
The number of moles of $F e S_{2} = \frac{500}{120} = 4.167$ moles.
$2$ moles of $F e S_{2}$ react with $8$ moles of $N a O H$ .
Hence, $4.167$ moles of $F e S_{2}$ will react with $\frac{8}{2} \times 4.167 = 16.67$ moles of $N a O H$ .
The volume of $6 M$ $N a O H$ solution required is $\frac{16.67}{6} = 2.778$ L $= 2778$ ml.

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