Đáp án:

 b) 2,4l

c) 10,08l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{0,8}}{2} = 0,4\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 11\\
1,5a + b = 0,4
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
\% {m_{Al}} = \dfrac{{0,2 \times 27}}{{11}} \times 100\%  = 49,1\% \\
\% {m_{Fe}} = 100 - 49,1 = 50,9\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,8\,mol\\
{n_{HCl}} = 0,8 \times 30\%  = 0,24\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = {n_{HCl}} = 0,24\,mol\\
{V_{{\rm{dd}}NaOH}} = \dfrac{{0,24}}{{0,1}} = 2,4l\\
c)\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
{n_{C{l_2}}} = 0,2 \times \dfrac{3}{2} + 0,1 \times \dfrac{3}{2} = 0,45\,mol\\
{V_{C{l_2}}} = 0,45 \times 22,4 = 10,08l
\end{array}\)