Em tham khảo nha :
\(\begin{array}{l}
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2mol\\
{n_{F{e_3}{O_4}}} = \dfrac{{17,4}}{{232}} = 0,075mol\\
8Al + 3F{e_3}{O_4} \to 4A{l_2}{O_3} + 9Fe\\
\frac{{0,2}}{8} = \dfrac{{0,075}}{3} \Rightarrow \\
\text{Gọi 8a là số mol của Al phản ứng }\\
{n_{A{l_{pu}}}} = 8a \Rightarrow {n_{Fe}} = \dfrac{9}{8}{n_{A{l_{pu}}}} = 9a\,mol\\
{n_{A{l_d}}} = 0,2 - 8a\,(mol)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}(1)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{5,376}}{{22,4}} = 0,24mol\\
\Rightarrow \dfrac{3}{2}{n_{A{l_d}}} + {n_{Fe}} = {n_{{H_2}}} = 0,24mol\\
\Rightarrow \dfrac{3}{2} \times (0,2 - 8a) + 9a = 0,24\\
\Rightarrow a = 0,02mol\\
H = \dfrac{{0,02 \times 8}}{{0,2}} \times 100\% = 80\%
\end{array}\)