What mass of `H_(2)SO_(4)` could be produced (in kg ) in the process given below if `5.52` kg of `FeS_(2)`is used
`FeS_(2)(s) + O_(2)(g) to Fe_(2)O_(3)(s)+SO_(2)(g)`
` SO_(2) (g) + O_(2)(g) to SO_(3)(g)`
`SO_(3)(g) + H_(2)SO_(4) (I) to H_(2)s_(2)O_(7)(I)`
`H_(2)S_(2)O_(7)(I) + H_(2)O(l) to H_(2)SO_(4)(sq)`
(At. wt. of `Fe = 56, S = 32`)

Explanation

Sulphuric acid contains one sulphur atom per molecule, and each Fes, molecule gives two atoms of sulphur.
Thus
`FeS_(2) rarr 2SO_(2) rarr 2SO_(2) rarr H_(2)SO_(4) rarr 2H_(2)SO_(4)`
Thus, 1 mol of `FeS_(2)` gives = 2 mol `H_(2)SO_(4) = 2 xx 98 g H_(2)SO_(4)`
Hence , 1.00 kg `FeS_(2) = 1000 g FeS_(2) = 1000/120 mol FeS_(2)`
` H_(2)SO_(4)` from 1000/120 mol `FeS_(2) = (2xx98xx1000)/120 g`
= 1.63 kg `H_(2)SO_(4)`