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 ]

Step-by-step solution

Number of moles of Cu NO 3 2 = 3 . 74 187 = 0 . 02 2 Cu NO 3 2 + 4 KI → Cu 2 I 2 ↓ + I 2 + 4 KNO 3 Number of moles of Cu 2 l 2 precipitated = 0 . 01 I 2 brown   solution + H 2 S → S ↓ + 2 HI Number of moles of S precipitated = 0 . 01 Mass of S precipitates = 0 . 01 × 32 g = 0 . 32   g