Đáp án:
\({V_{{\rm{dd}}AgN{O_3}}} = 3,5l\)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{{H_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
\left\{ \begin{array}{l}
27a + 56b = 13,9\\
1,5a + b = 0,35
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
AlC{l_3} + 3AgN{O_3} \to 3AgCl + Al{(N{O_3})_3}\\
FeC{l_2} + 2AgN{O_3} \to 2AgCl + Fe{(N{O_3})_2}\\
{n_{AgN{O_3}}} = 0,1 \times 3 + 0,2 \times 2 = 0,7\,mol\\
{V_{{\rm{dd}}AgN{O_3}}} = \dfrac{{0,7}}{{0,2}} = 3,5l
\end{array}\)