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  4. The minimum quantity of H2S needed to precipitate 64.5 g of Cu2+ will be nearly.

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Q. The minimum quantity of $H_2S$ needed to precipitate $64.5\, g$ of $Cu^{2+}$ will be nearly.

Some Basic Concepts of Chemistry

Solution:

Meq. of $H_{2}S =$ Meq. of $Cu^{2+}$
$ \therefore \frac{\,{}^{w}H_{2}S}{342}\times1000 = \frac{63.5}{63.52}\times1000$
$\therefore \,{}^{w}H_{2}S = 34 \,g$