Zinc blende is roasted by the reaction 2ZnS + 3O2 -> 2ZnO + 2SO2. Taking Zn = 65, S = 32, O = 16 and molar volume 22.4 L/mol at STP, the mass of O2 needed and the volume of SO2 formed at STP when 4.85 kg of pure ZnS is roasted completely are

JEE Main levelSingle correct optionDifficulty 2 of 3mt022

Zinc blende is roasted by the reaction $2 ZnS + 3 O X_{2} 2 ZnO + 2 SO X_{2}$ . Taking $Zn = 65$ , $S = 32$ , $O = 16$ and molar volume $22.4 L/mol$ at STP, the mass of $O X_{2}$ needed and the volume of $SO X_{2}$ formed at STP when $4.85 kg$ of pure $ZnS$ is roasted completely are

The molar mass of

ZnS

is 97, so

4850 g

is 50 mol. From the equation each mole of ZnS needs 1.5 mol

O X_{2}

, i.e. 75 mol

= 2400 g = 2.40 kg

, and gives 50 mol

SO X_{2}

= 50 \times 22.4 = 1120 L