In the following reaction, which choice has value twice that of the equivalent mass of the oxidising agentSO2+2H2S⟶3S+2H2O

In the following reaction, which choice has value twice that of the equivalent mass of the oxidising agent
${SO}_{2} + 2 H_{2} S ⟶ 3 S + 2 H_{2} O$

A
64
B
32
C
16
D
48

Solution:

$\begin{array}{l} {SO}_{2} + 2 H_{2} S \to 3 S + 2 H_{2} O \\ + 4 - 2 0 \\ EW = \frac{M}{4} = \frac{64}{4} = 16; Twice 16 \times 2 = 32 . \end{array}$

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