Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,15mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
\% Fe = \dfrac{{8,4}}{{12}} \times 100\% = 70\% \\
\% Cu = 100 - 70 = 30\% \\
c)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,3mol\\
{m_{HCl}} = 0,3 \times 36,5 = 10,95g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{10,95 \times 100}}{{10}} = 109,5g\\
d)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,15mol\\
{m_{FeC{l_2}}} = 0,15 \times 127 = 19,05g\\
C{\% _{FeC{l_2}}} = \dfrac{{19,05}}{{8,4 + 109,5 - 0,15 \times 2}} \times 100\% = 16,2\%
\end{array}\)
