JEE Advance - Chemistry (2022 - Paper 1 Online - No. 4)
The treatment of an aqueous solution of $3.74 \mathrm{~g}$ of $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ with excess KI results in a brown solution along with the formation of a precipitate. Passing $\mathrm{H}_{2} \mathrm{~S}$ through this brown solution gives another precipitate $\mathbf{X}$. The amount of $\mathbf{X}$ (in $g$ ) is ___________.
[Given: Atomic mass of $\mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \mathrm{~K}=39, \mathrm{Cu}=63, \mathrm{I}=127$ ]
Explanation
Number of moles of $$Cu{(N{O_3})_2} = {{3.74} \over {187}} = 0.02$$
$$\mathop {2Cu{{(N{O_3})}_2}}\limits_{0.02} + 4KI \to C{u_2}{I_2} \downarrow + \mathop {{I_2}}\limits_{0.01} + 4KN{O_3}$$
$$\mathop {{I_2}}\limits_{0.01} + KI \to \mathop {K{I_3}}\limits_{0.01\,(Brown\,solution)} $$
$$\mathop {K{I_3}}\limits_{0.01} + {H_2}S \to KI + \mathop S\limits_{0.01\,(X)} \downarrow + 2HI$$
Number of moles of sulphur precipitated (X) = 0.01
Mass of sulphur precipitates (X) = 0.01 $$\times$$ 32 = 0.32 gm
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