The ratio of amounts of $\mathrm{H}_{2} \mathrm{~S}$ needed to precipitate all the metal ions from $100…

The ratio of amounts of $\mathrm{H}_{2} \mathrm{~S}$ needed to precipitate all the metal ions from $100 \mathrm{ml}$ of $1 \mathrm{M}$ $\mathrm{AgNO}_{3}$ and $100 \mathrm{ml}$ of $1 \mathrm{M} \mathrm{CuSO}_{4}$ will be
  1. $1: 1$
  2. $1: 2$
  3. $2: 1$
  4. None of these

Solution

$\mathrm{AgNO}_{3} \equiv 2 \mathrm{Ag}^{+}+\underset{\left(H_{2} \mathrm{~S}ight)}{\mathrm{S}^{2-}} ightarrow \mathrm{Ag}_{2} \mathrm{~S}$
$\because 2$ mole $ightarrow 1$ mole $\quad[100 \times 1=100 millimole]$
$\therefore 100$ millimole $ightarrow 50$ millimole $H_{2} S$ required
$\mathrm{CuSO}_{4} \equiv \mathrm{Cu}^{2+}+\underset{\left(H_{2} \mathrm{~S}ight)}{\mathrm{S}^{2-}} ightarrow \mathrm{Cu}\mathrm{~S}$
1 mole $ightarrow 1$ mole $(100 \times 1=100$ millimole $)$
100 milimole $ightarrow 100$ milimole $\mathrm{H}_{2} \mathrm{~S}$ required
Ratio $\frac{50}{100}=\frac{1}{2}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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