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Q.

In the following conversion of sulphide of phosphorous: $P_{4} S_{3} ⟶ P_{2} O_{5} + {SO}_{2}$

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a

$\frac{M}{14}$

b

$\frac{M}{38}$

c

$\frac{M}{18}$

d

$\frac{M}{32}$

answer is C.

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Detailed Solution

Equivalent weight (E) ,: $P_{4} S_{3} = \frac{M}{32}$

$16 H_{2} O + P_{4} S_{3} ⟶ 2 P_{2} O_{5} + 3 S O_{2} + 32 H^{+} + 32 e^{-}$

$P_{4}^{1} . 5 S_{3}^{- 2} ⟶ (2 P_{2}) ┬ (14 e^{-}) O_{5} + 3 S O_{2}^{+ 4}$

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