Đáp án:

 b) 40,68%

c) 0,22 l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
Cu + C{l_2} \to CuC{l_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,2\,mol\\
{n_{FeC{l_3}}} = {n_{Fe}} = 0,2\,mol\\
{m_{CuC{l_2}}} = 48,7 - 0,2 \times 162,5 = 16,2g\\
{n_{CuC{l_2}}} = \dfrac{{16,2}}{{135}} = 0,12\,mol\\
{n_{Cu}} = {n_{CuC{l_2}}} = 0,12\,mol\\
m = {m_{Cu}} + {m_{Fe}} = 0,12 \times 64 + 0,2 \times 56 = 18,88g\\
\% {m_{Cu}} = \dfrac{{0,12 \times 64}}{{18,88}} \times 100\%  = 40,68\% \\
\% {m_{Fe}} = 100 - 40,68 = 59,32\% \\
c)\\
{n_{HCl}} = 0,2 \times 2 \times 110\%  = 0,44\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{0,44}}{2} = 0,22l
\end{array}\)