JEE Advanced levelSingle correct optionDifficulty 3 of 3mt035
A heavy, lead-grey sulphide ore X is concentrated by froth flotation. It is roasted in a limited supply of air to form some oxide together with a pungent gas that turns acidified dichromate green; the air is then cut off and the temperature raised so that the sulphide and oxide react to give the metal and more of the same gas (PbS+2PbO3Pb+SOX2 if X is the lead ore). Taking Pb=207, S=32, O=16 and 22.4L/mol at STP, complete conversion of 1.195kg of pure X gives metal of mass and, in the roasting stage alone, gas of volume at STP
A1.035 kg and 74.7 L
B1.035 kg and 37.3 L
C0.690 kg and 74.7 L
D1.035 kg and 112.0 L
Answer: A
X is galena, PbS (molar mass 239), so 1195g is 5.00 mol. In the roasting stage 2PbS+3OX22PbO+2SOX2, and for the self-reduction PbS+2PbO3Pb+SOX2 we need 2 mol PbO per mol PbS, so 2/3 of the PbS is roasted (3.33 mol, giving 3.33 mol SOX2=74.7L) and 1/3 is left to self-reduce. Overall each mol PbS gives 1 mol Pb, so the metal is 5.00×207=1.035kg (the self-reduction stage adds a further 37.3 L of SOX2).