In a quality control analysis for sulphur impurity 0.56g steel sample was burnt in a stream of oxygen and sulpphur was converted into SO_(2) gas. The SO_(2) was then oxidized to sulphate by using H_(2)O_(2) solution to which had been added 30mL of 0.04M NaOH. the equation for reaction is: SO_(2(g)) +H_(2)O_(2(aq)) +2OH_((aq))^(-) rarr SO_(4(aq))^(-2) +2H_(2)O_((1)) 22.48 mL of 0.024M HCI was required to neutralize the base remaining after oxidation reaction. Calculate % of sulphur in given sample.

In a quality control analysis for sulphur impurity 0.56g steel sample

1.	In a quality control analysis for sulphur impurity 0.56g steel sample was burnt in a stream of oxygen and sulpphur was converted into SO_(2) gas. The SO_(2) was then oxidized to sulphate by using H_(2)O_(2) solution to which had been added 30mL of 0.04M NaOH. the equation for reaction is: SO_(2(g)) +H_(2)O_(2(aq)) +2OH_((aq))^(-) rarr SO_(4(aq))^(-2) +2H_(2)O_((1)) 22.48 mL of 0.024M HCI was required to neutralize the base remaining after oxidation reaction. Calculate % of sulphur in given sample.
Answer» `3.6%` `1.875%` `9%` `4.5%` SOLUTION :Meq of alkali ADDED `=30 xx 0.04 = 1.2` Meq. Of alkali left `=22.48 xx 0.024 = 0.54` `:.` Meq. of alkali for `SO_(2)` and `H_(2)O_(2) = 1.2 - 0.54 = 0.66 =` M.eqts of alkali 1 moles of alkai `-=1` moles of `SO_(2)` `:.` WEIGHT of `S = (0.66)/(2) xx (32)/(1000) = 0.0105g` `:. %` of `S = (0.0105)/(0.56) xx 100 = 1.875%`