Đáp án:
9,96 g
Giải thích các bước giải:
\(\begin{array}{l}
TN1:\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
TN2:\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
n{H_2} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
nS{O_2} = \dfrac{{8,064}}{{22,4}} = 0,36\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
1,5a + b = 0,3\\
1,5a + 1,5b = 0,36\\
\Rightarrow a = b = 0,12\\
m = 0,12 \times 27 + 0,12 \times 56 = 9,96g
\end{array}\)
