Đặt công thức oleum là \(H_2SO_4.nSO_3\)
Số mol NaOH là 0,2 mol
\(\begin{array}{l}
{H_2}S{O_4}.nS{O_3} + n{H_2}O \to \left( {n + 1} \right){H_2}S{O_4}\\
\dfrac{{0,1}}{{n + 1}}mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,mol\\
{H_2}S{O_4}\, + \,2NaOH\, \to N{a_2}S{O_4} + 2{H_2}O\\
0,1\,mol\,\,\,\,\,\,\,\,0,2\,mol
\end{array}\)
\(\begin{array}{l}
{M_{{H_2}S{O_4}.nS{O_3}}} = \dfrac{{8,45}}{{0,1}}\left( {n + 1} \right) = 98 + 80n\\
\Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,84,5n + 84,5 = 98 + 80n\\
\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n = 3
\end{array}\)
vậy công thức của A : \(H_2SO_4.3SO_3\)
