In a quality control analysis for sulphur impurity `5.6g` steel sample was burnt in a stream of oxygen and sulphur was converted into `SO_(2)` gas. The `SO_(2)` was then oxidized to sulphate by using `H_(2)O_(2)` solution to which has been added `30 mL` of `0.04M NaOH`. The equation of the reaction is:
`SO_(2(g)) +H_(2)O_((aq.))+2OH_((aq.))^(-)to SO_(4(aq.))^(2-) +2H_(2)O_((l))`
`22.48mL` of `0.024M HCI` was required to neutralize the base remaining after oxidation reaction. Calculate `%` sulphur in given sample.

Answer

`0.1875%`