Đáp án:
D
Giải thích các bước giải:
\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
MgS{O_4} + 2NaOH \to 2N{a_2}S{O_4} + Mg{(OH)_2}\\
A{l_2}{(S{O_4})_3} + 6NaOH \to 2Al{(OH)_3} + 3N{a_2}S{O_4}\\
FeS{O_4} + 2NaOH \to Fe{(OH)_2} + N{a_2}S{O_4}\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O\\
4Fe{(OH)_2} + {O_2} + 2{H_2}O \to 4Fe{(OH)_3}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
Mg{(OH)_2} \to MgO + {H_2}O\\
n{H_2} = \frac{{2,24}}{{22,4}} = 0,1\,mol\\
hh:Mg(a\,mol),Al(b\,mol),Fe(c\,mol)\\
24a + 27b + 56c = 2,43\\
a + 1,5b + c = 0,1\\
40a + 80c = 1,6\\
= > a = 0,01;b = 0,05;c = 0,015\\
\% mAl = \frac{{0,05 \times 27}}{{2,43}} \times 100\% = 55,5\%
\end{array}\)