6.0 g of a steel containing sulphur as an impurity was burnt in excess of oxygen, where sulphur is oxidised to `SO_2`. The `SO_2` evolved was oxidised to `SO_4^(2-)` ions by the action of `H_2O_2` solution in the presence of 30 mL solution of 0.04 M NaOH. 25 " mL of " 0.02 M HCl was required to neutralise the excess of NaOH after the above oxidation. Calculate the percentage of sulphur in the given sample of steel (Atomic mass of S is 32).

Explanation

6 g steel has S as impurity.
`S+O_2toSO_2underset(NaOH)overset(H_2O_2)toSO_4^(2-)`
`(10mL,0.04M)` of excess `NaOH-=25mL(0.02M)HCl`
`SO_2+H_2O_2+2NaOHtoNa_2SO_4+2H_2O`
`1 m mol NaOH-=(1)/(2)mmol SO_2`
[Also, 1 m mol `SO_2-=1mmolS`]
Excess m" Eq of "`NaOH=m" Eq of "HCl`
`=25xx0.02=0.5`
`impliesm" Eq of "NaOH used =30xx0.04-0.5`
`=1.2-0.5=0.7`
`impliesm" mol of "NaOH used =0.7`
From stoichiometry: m" mol of "`SO_2=(0.7)/(2)=m" mol of "S`
`implies0.35=(weight)/(32)xx1000impliesweight=0.0112g`
`%S=(0.0112)/(6)xx100=0.186%`