JEE Advanced 2022 · Paper 1 · Chemistry Q4

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ChemistryNumerical+3Question 40 of 54

The treatment of an aqueous solution of 3.74 g of $\ce{Cu(NO3)2}$ with excess KI results in a brown solution along with the formation of a precipitate. Passing $\ce{H2S}$ through this brown solution gives another precipitate X. The amount of X (in g) is _____.

[Given: Atomic mass of H = 1, N = 14, O = 16, S = 32, K = 39, Cu = 63, I = 127]

Your answer

Answer and worked solution

Official answer: 0.31

Moles of copper. $\ce{Cu(NO3)2}$ has $M=63+2(62)=187$ , so 3.74 g is 0.02 mol.

With KI.
$\ce{2Cu^2+ + 4I- -> 2CuI v + I2}$
This gives 0.01 mol $\ce{I2}$ , which stays in solution as brown $\ce{I3-}$ .

With $\ce{H2S}$ .
$\ce{I2 + H2S -> S v + 2HI}$
X is sulphur: 0.01 mol $=0.32$ g.

Topics:Redox titrations Transition element properties Stoichiometry

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