The treatment of an aqueous solution of 3.74 g of Cu ( NO 3 ) 2 with excess KI results in a brown solution along with the formation of a precipitate. Passing H 2 S through this brown solution gives another precipitate X . The amount of X (in g ) is .[0pt] [Given: Atomic mass of H =1, N =14, O =16, S =32, K =39, Cu =63, I =127 ]

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Advanced 2022 — Paper 1 — Question 38

The treatment of an aqueous solution of 3.74 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 $\qquad$ .[0pt]

[Given: Atomic mass of $\mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \mathrm{~K}=39, \mathrm{Cu}=63, \mathrm{I}=127$ ]

Numerical answer — enter this value.

$2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+4 \mathrm{KI} \longrightarrow 2 \mathrm{CuI} \downarrow+\mathrm{I}_{2}+4 \mathrm{KNO}_{3}$

$\mathrm{I}_{2}+\mathrm{KI} \longrightarrow \mathrm{KI}_{3}$

$2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+5 \mathrm{KI} \longrightarrow 2 \mathrm{CuI} \downarrow+\mathrm{KI}_{3}+4 \mathrm{KNO}_{3}$

Mole $=\frac{3.74}{187}=0.02 \quad \frac{0.02}{2}=0.01$

$\underset{\mathrm{KI}_{3}}{ }+\mathrm{H}_{2} \mathrm{~S} \longrightarrow \mathrm{~S} \downarrow+2 \mathrm{HI}+\mathrm{KI}$

0.01 mol 0.01 mol $(\mathrm{X})$ Mass of ' S ' $=0.01 \times 32=0.32$

Answer key and solution verified before publishing.

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