Chalcopyrite is roasted so that its iron is converted to FeO (the first step being 2CuFeS2 + O2 -> Cu2S + 2FeS + SO2 followed by 2FeS + 3O2 -> 2FeO + 2SO2), and then smelted with silica. Taking Cu = 63.5, Fe = 56, S = 32, Si = 28, O = 16, what mass of SiO2 is needed to slag all iron from 367 g of pure CuFeS2?

JEE Advanced levelSingle correct optionDifficulty 3 of 3mt144

Chalcopyrite is roasted so that its iron is converted to $FeO$ (the first step being $2 CuFeS X_{2} + O X_{2} Cu X_{2} S + 2 FeS + SO X_{2}$ followed by $2 FeS + 3 O X_{2} 2 FeO + 2 SO X_{2}$ ), and then smelted with silica. Taking Cu = 63.5, Fe = 56, S = 32, Si = 28, O = 16, what mass of $SiO X_{2}$ is needed to slag all iron from $367 g$ of pure $CuFeS X_{2}$ ?

The molar mass of

CuFeS X_{2}

63.5 + 56 + 64 = 183.5

, so

367 g

2.00 mol

. Each mole gives one mole of

FeO

and

FeO + SiO X_{2} FeSiO X_{3}

needs one mole of silica. Silica needed is

2.00 \times 60 = 120 g