Chemistry - Concentration Terms Question with Solution | TestHub

$2\text{N}\text{a}\text{O}\text{H}+{\text{H}}_2\text{S}{\text{O}}_4\to \text{N}{\text{a}}_2\text{S}{\text{O}}_4+2{\text{H}}_2\text{O}$
$2\text{N}\text{a}\text{O}\text{H}+\text{S}{\text{O}}_3\to \text{N}{\text{a}}_2\text{S}{\text{O}}_4+2{\text{H}}_2\text{O}$
${\mathrm{M}}_{\mathrm{e}\mathrm{q}}$of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4+{\mathrm{M}}_{\mathrm{e}\mathrm{q}}$of$\mathrm{S}{\mathrm{O}}_3={\mathrm{M}}_{\mathrm{e}\mathrm{q}}$of NaOH
$\frac{(0.5-\mathrm{a})}{49}\times 1000+\frac{\mathrm{a}}{80/2}\times 1000=26.7\times 0.4$
So$\mathrm{a}=0.103$
So$\mathrm{\% }\mathrm{o}\mathrm{f }\mathrm{S}{\mathrm{O}}_3=\frac{0.103}{0.5}\times 100=20.6\mathrm{\%}$.