The equivalents of $H_{2} O_{2}$ left after reacting with $S n^{2 +}$ is:

0.1

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0.2

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0.3

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0.4

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Solution

The correct option is B $0.2$
Moles of $H_{2} O_{2}$ initially present $= 136 \times \frac{10}{100} \times \frac{1}{34}$ $= 0.4$ mol.
Moles of $H_{2} O_{2}$ left $=$ Mol of $H_{2} O_{2}$ initially present $-$ Mol of $H_{2} O_{2}$ reduced by $S n^{2 +}$ $= 0.4 - 0.3 = 0.1$ mol
Equivalents of $H_{2} O_{2}$ left $= 0.1 \times 2 = 0.2$ eq. per $100$ mL.

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