Đáp án:

\(\begin{array}{l}
a)\\
{m_A} = 21,2g\\
b)\\
{C_M}FeC{l_2} = 0,3M\\
{C_\% }CuS{O_4} = 10,67\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,15\,mol\\
{n_{S{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{Cu}} = {n_{S{O_2}}} = 0,2\,mol\\
{m_A} = {m_{Fe}} + {m_{Cu}} = 0,15 \times 56 + 0,2 \times 64 = 21,2g\\
b)\\
{V_{{\rm{dd}}X}} = {V_{{\rm{dd}}HCl}} = 0,5l\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,15\,mol\\
{C_M}FeC{l_2} = \dfrac{{0,15}}{{0,5}} = 0,3M\\
{m_{{\rm{dd}}spu}} = 0,2 \times 64 + 300 - 0,2 \times 64 = 300g\\
{n_{CuS{O_4}}} = {n_{Cu}} = 0,2mol\\
{C_\% }CuS{O_4} = \dfrac{{0,2 \times 160}}{{300}} \times 100\%  = 10,67\% 
\end{array}\)