Em tham khảo nha :
\(\begin{array}{l}
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{A{l_2}{O_3}}} = \dfrac{{10,2}}{{102}} = 0,1mol\\
{n_{MgO}} = \dfrac{4}{{40}} = 0,1mol\\
{n_{NaOH}} = 0,2 \times 0,1 = 0,02mol\\
{n_{{H_2}S{O_4}}} = 3{n_{A{l_2}{O_3}}} + {n_{MgO}} + \dfrac{{{n_{NaOH}}}}{2} = 0,41mol\\
{m_{{H_2}S{O_4}}} = 0,41 \times 98 = 40,18g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{40,18}}{{245}} = 16,4\%
\end{array}\)