What mass of `H_(2)SO_(4)` could be produced in the process given below if 1.00 kg of `FeS_(2)` is used? The unbalanced equations for the process are:
`FeS_(2)(s) + O_(2)(g) rarr Fe_(2)O_(3)(s) + SO_(2)(g)`
`SO_(2)(g) + O_(2)(g) rarr SO_(3)(g)`
`SO_(3)(g) + H_(2)SO_(4)(l)rarr H_(2)SO(4)(l)`
` H_(2)S_(2)O_(7)(l) rarr H_(2)SO_(4) (aq) ` (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)`