Đáp án:

10,08l

Giải thích các bước giải:

\(\begin{array}{l}
{n_{{H_2}({P_1})}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{{H_2}({P_2})}} = \dfrac{{7,84}}{{22,4}} = 0,35\,mol\\
{n_{{H_2}({P_1})}} < {n_{{H_2}({P_2})}} \Rightarrow \text{ Phần 1 Al còn dư } \\
hh:Na(a\,mol),Al(b\,mol),Fe(c\,mol)\\
 \Rightarrow 23a + 27b + 56c = 39,9(1)\\
{P_1}:\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
2Al + 2NaOH + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}\\
 \Rightarrow \dfrac{a}{3} \times \dfrac{1}{2} + \dfrac{a}{3} \times \dfrac{3}{2} = 0,2 \Rightarrow a = 0,3\\
{P_2}:\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
2Al + 2NaOH + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}\\
\dfrac{a}{3} \times \dfrac{1}{2} + \dfrac{b}{3} \times \dfrac{3}{2} = 0,35 \Rightarrow b = 0,6\\
(1) \Rightarrow c = 16,8g\\
{P_3}:\\
2Na + 2HCl \to 2NaCl + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{0,3}}{3} \times \dfrac{1}{2} + \dfrac{{0,6}}{3} \times \dfrac{3}{2} + \dfrac{{16,8}}{{56}} \times \dfrac{1}{3} = 0,45\,mol\\
V = 0,45 \times 22,4 = 10,08l
\end{array}\)