Em tham khảo nha :

\(\begin{array}{l}
1)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,15mol\\
{m_{Mg}} = 0,15 \times 24 = 3,6g\\
{m_{Cu}} = 10 - 3,6 = 6,4g\\
2)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35mol\\
hh:Fe(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,35\\
56a + 65b = 21,4
\end{array} \right.\\
 \Rightarrow a = 0,15;b = 0,2\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
\% Fe = \dfrac{{8,4}}{{21,4}} \times 100\%  = 39,25\% \\
\% Zn = 100 - 39,25 = 60,75\% \\
b)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,15mol\\
{m_{FeC{l_2}}} = 0,15 \times 127 = 19,05g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,2mol\\
{m_{ZnC{l_2}}} = 0,2 \times 136 = 27,2g\\
{m_m} = 19,05 + 27,2 = 46,25g
\end{array}\)

Đáp án:

 a) %mFe=39,25% -> %mZn=60,75%

b)

mFeCl2=19,05 gam

 mZnCl2=27,2 gam

Giải thích các bước giải:

 Gọi số mol Fe là x; Zn là y ->56x+65y=21,4

Fe + 2HCl -> FeCl2 + H2

Zn + 2HCl -> ZnCl2 + H2

-> nH2=nFe+nZn=x+y=7,84/22,4=0,35 mol

Giải được:  x=0,15; y=0,2

-> mFe=0,15.56=8,4 gam

->%mFe=8,4/21,4=39,25% -> %mZn=60,75%

Ta có : nFeCl2=nFe=0,15 mol -> mFeCl2=0,15.(56+35,5.2)=19,05 gam

nZnCl2=nZn=0,2 mol -> mZnCl2=0,2.(65+35,5.2)=27,2 gam