JEE Advanced levelSingle correct optionDifficulty 3 of 3mt151
A metal M occurs as a sulphide ore concentrated by froth flotation. The ore is roasted, the oxide is heated with coke near 1673K and M leaves the furnace as vapour (boiling point 1180K). The metal also precipitates gold from Na[Au(CN)X2] solution. Roasting 48.5g of the pure sulphide (M = 65, S = 32) at STP gives what volume of SOX2, and what mass of M is obtained on complete reduction of the oxide?
A22.4L of SOX2 and 32.5g of Zn
B11.2L of SOX2 and 65g of Zn
C11.2L of SOX2 and 32.5g of Zn
D16.8L of SOX2 and 16.25g of Zn
Answer: C
A volatile metal obtained by coke reduction at 1673K that displaces gold is zinc, so the ore is ZnS (molar mass 97). 48.5g is 0.500mol. From 2ZnS+3OX22ZnO+2SOX2, 0.500mol gives 0.500molSOX2, i.e. 11.2L, and 0.500mol ZnO, which gives 0.500×65=32.5g Zn.