Đáp án:
5,6g
Giải thích các bước giải:
\(\begin{array}{l}
hh:Na(a\,mol),Fe(b\,mol),Al(c\,mol)\\
2Na + 2HCl \to 2NaCl + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35\,mol \Rightarrow 0,5a + b + 1,5b = 0,35(1)\\
2Na + C{l_2} \to 2NaCl\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
{n_{C{l_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol \Rightarrow 0,5a + 1,5b + 1,5c = 0,4(2)\\
(2) - (1) \Rightarrow 0,5b = 0,4 - 0,35 \Rightarrow b = 0,1\\
{m_{Fe}} = 0,1 \times 56 = 5,6g
\end{array}\)