Chemistry - Concentration Terms Question with Solution | TestHub

Total moles of $(H_{2}S{O}_4+S{O}_3+S{O}_2)$$=\frac{MolesofNaOH}{2}=\frac{18\times 2}{2}=18moles$.

Moles of $(H_{2}S{O}_4+S{O}_3)$$=18-5(molesofS{O}_2)=13moles$.

Solving for mass, $98\times n_{H_{2}S{O}_4}+80\times (13-n_{H_{2}S{O}_4})=1450-(5\times 64)$ gives $n_{H_{2}S{O}_4}=5moles$.

Combined $S{O}_3$ is chemically bound inside $H_{2}S{O}_4$, so Moles of combined $S{O}_3=n_{H_{2}S{O}_4}=5moles$.

Percentage of combined $S{O}_3$$=\frac{5\times 80}{1450}\times 100=27.59\%$ (Option C).