Chemistry - Concentration Terms Question with Solution | TestHub

Total moles of NaOH used for neutralization is $18L\times 2.00M=36moles$.

This NaOH reacts completely with both H₂SO₄ and free SO₃ (which forms H₂SO₄ with water), requiring $2$ moles of NaOH per mole of H₂SO₄ or SO₃. Thus, total moles of $(H_{2}S{O}_4+S{O}_3)=\frac{36}{2}=18moles$. Let the mass of free SO₃ be $x$ and H₂SO₄ be $y$; then $x+y=1450g-(5moles\times 64g/molofS{O}_2)=1130g$. Solving the system $\frac{x}{80}+\frac{y}{98}=18$ and $x+y=1130$ yields $x=640g$ of free SO₃.

The mass percentage of free SO₃ is $\frac{640g}{1450g}\times 100\%=44.14\%$ (Option A).