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

Question

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 [Given: Atomic mass of H =1, N =14, O =16, S =32, K =39, Cu =63, I =127 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

underset0.022 Cu ( NO 3)2+5 KI longrightarrow Cu 2 I 2+ underset0.01KI 3+4 KNO 3 underset0.01KI 3+ H 2 S longrightarrow underset0.01S downarrow+ KI +2 HI n S =0.01 mole weight of sulphur =32 × 0.01=0.32 gm

0.022Cu(NO3​)2​​+5KI⟶Cu2​I2​+0.01KI3​​+4KNO3​ 0.01KI3​​+H2​S⟶0.01S↓​+KI+2HI nS​=0.01 mole weight of sulphur =32×0.01=0.32gm

$0.02 2 C u (N O_{3})_{2} + 5 K I ⟶ C u_{2} I_{2} + 0.01 K I_{3} + 4 K N O_{3}$
$0.01 K I_{3} + H_{2} S ⟶ 0.01 S ↓ + K I + 2 H I$
$n_{S} = 0.01$ mole
weight of sulphur $= 32 \times 0.01 = 0.32 g m$