$n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)\\ PTHH:\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ \Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ \Rightarrow m_{Al_2O_3}=15,6-5,4=10,2(g)\\ b,n_{Al_2O_3}=\dfrac{10,2}{102}=0,1(mol)\\ \Rightarrow \Sigma n_{Al_2(SO_4)_3}=n_{Al_2O_3}+n_{Al}=0,3(mol)\\ m_{\text{dd sau p/ứ}}=m_X+m_{dd_{H_2SO_4}}-m_{H_2}=15,6+392-0,3.2=407(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,3.342}{407}.100\%\approx 25,21\%$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{{7,437}}{{24,79}} = 0,3mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
{m_{A{l_2}{O_3}}} = 15,6 = 5,4 = 10,2g\\
b)\\
{n_{A{l_2}{O_3}}} = \dfrac{{10,2}}{{102}} = 0,1mol\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} + {n_{A{l_2}{O_3}}} = 0,2mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,2 \times 342 = 68,4g\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{68,4}}{{392 + 15,6 - 0,3 \times 2}} \times 100\%  = 16,8\% 
\end{array}\)