Em tham khảo nha :
\(\begin{array}{l}
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{A{l_2}{O_3}}} = \dfrac{{10,2}}{{102}} = 0,1mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{39,2}}{{98}} = 0,4mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,4}}{3} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}S{O_4}d}} = 0,4 - 3 \times 0,1 = 0,1mol\\
{m_{{H_2}S{O_4}d}} = 0,1 \times 98 = 9,8g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = {n_{A{l_2}{O_3}}} = 0,1mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,1 \times 342 = 34,2g
\end{array}\)