\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.15.....0.3.......................0.15\)

\(m_{Mg}=0.15\cdot24=3.6\left(g\right)\)

\(m_{Cu}=10-3.6=6.4\left(g\right)\)

\(\%Mg=\dfrac{3.6}{10}\cdot100\%36\%\)

\(\%Cu=64\%\)

\(V_{dd_{HCl}}=\dfrac{0.3}{2}=0.15\left(l\right)\)

a, Ta có: 24n_Mg + 56n_Fe = 9,2 (g) (1)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

BT e, có: 2n_Mg + 2n_Fe = 2n_H2 = 0,5 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,15\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\%\approx39,13\%\\\%m_{Fe}\approx60,87\%\end{matrix}\right.\)

b, BTNT H, có: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)

PTHH: 

\(Zn+H_2SO_4--->ZnSO_4+H_2\)

\(Cu+H_2SO_4--\times-->\)

a. Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)

\(\Rightarrow\%_{m_{Zn}}=\dfrac{6,5}{19,3}.100\%=33,7\%\)

\(\%_{m_{Cu}}=100\%-33,7\%=66,3\%\)

b. Theo PT: \(n_{H_2SO_4}=n_{Zn}=0,1\left(mol\right)\)

Đổi 200ml = 0,2 lít

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5M\)

c. Ta có: \(V_{dd_{ZnSO_4}}=V_{dd_{H_2SO_4}}=0,2\left(lít\right)\)

Theo PT: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{ZnSO_4}}=\dfrac{0,1}{0,2}=0,5M\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(n_{Mg}=n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(\Rightarrow m_{Mg}=0.25\cdot24=6\left(g\right)\)

\(m_{Mg}>m_{hh}=5.6\left(g\right)\)

=> Đề sai 

nH2= 1,12/22,4=0,05(mol)

PTHH: Fe +2 HCl -> FeCl2 + H2

0,05_______0,1__0,05___0,05(mol)

a) mFe=0,05.56=2,8(g)

=>%mFe=(2,8/10).100=28% => %mCu=100%- 28%= 72%

b) mHCl=0,1.36,5=3,65(g)

=> C%ddHCl= (3,65/200).100= 1,825%

nH2= 1,12/22,4=0,05(mol)

PTHH: Fe +2 HCl -> FeCl2 + H2

0,05_______0,1__0,05___0,05(mol)

a) mFe=0,05.56=2,8(g)

=>%mFe=(2,8/10).100=28% => %mCu=100%- 28%= 72%

b) mHCl=0,1.36,5=3,65(g)

=> C%ddHCl= (3,65/200).100= 1,825%