\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

___________________0,3<----0,3____(mol)

=> m_MgCl2 = 0,3.95 = 28,5 (g)

Chất rắn không tan là Ag

\(\Rightarrow m_{Ag} = 4,6(gam)\)

\(n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{Al} = \dfrac{2}{3}n_{H_2} = 0,2(mol)\\ m_{Al} = 0,2.27 = 5,4(gam)\)

a) PTHH: 2Al + 6 HCl -> 2 AlCl3 + 3 H2

x___________3x______________1,5x(mol)

Fe +2 HCl -> FeCl2 + H2

y___2y____y______y(mol)

b) Ta có: m(rắn)= mCu=0,4(g)

=> m(Al, Fe)=1,5-mCu=1,5-0,4=1,1(g)

nH2= 0,04(mol)

Ta lập hpt:

\(\left\{{}\begin{matrix}27x+56y=1,1\\1,5x+y=0,04\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,02\\y=0,01\end{matrix}\right.\)

=> mAl=27.0,02=0,54(g)

mFe=56.0,01=0,56(g)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl -->ZnCl₂ + H₂

____0,2<----------------------0,2

=> m_Zn = 0,2.65 = 13 (g)

m_Cu = m_{rắn không tan} = 19,5 (g)

\(\left\{{}\begin{matrix}\%Zn=\dfrac{13}{13+19,5}.100\%=40\%\\\%Cu=\dfrac{19,5}{13+19,5}.100\%=60\%\end{matrix}\right.\)

`n_(H_2)=4,48/22,4=0,2 (mol)`

Ta có PTHH: `Zn+2HCl --> ZnCl_2 +H_2`

Theo PT:       `1`--------------------------------`1`

Theo đề:       `0,2`------------------------------`0,2`

`m_(Zn)=0,2.65=13(g)`

Vì `Cu` không phản ứng với `HCl` nên `m_(chất rắn không tan)=m_(Cu)=19,5(gam)`

`%Zn=13/(13+19,5) .100%=40%`

`%Cu=100%-40%=60%`

a)

\(n_{H_2\left(1\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂ (1)

             0,6<----------------------0,3

=> m_Na = 0,6.23 = 13,8 (g)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,1<-0,2

=> m_Fe = 0,1.56 = 5,6 (g)

m_Cu = 10 (g)

\(\left\{{}\begin{matrix}\%Na=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%Fe=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%Cu=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)

PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

             \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=\dfrac{17,4}{\dfrac{0,3}{y}}=58y\left(g/mol\right)\)

=> 56x = 42y

=> \(\dfrac{x}{y}=\dfrac{3}{4}\) => CTHH: Fe₃O₄

a)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

            0,6<----------------------0,3

             Fe + 2HCl --> FeCl₂ + H₂

            0,1<--0,2

=> \(\left\{{}\begin{matrix}m_{Na}=0,6.23=13,8\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Cu}=10\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%m_{Fe}=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%m_{Cu}=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)
PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

            \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=56x+16y=\dfrac{17,4}{\dfrac{0,3}{y}}\left(g/mol\right)\)

=> \(\dfrac{x}{y}=\dfrac{3}{4}\)

=> CTHH: Fe₃O₄

\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 16,8 - 6,4 = 10,4(1)\\ Fe + 2HCl \to FeCl_2 + H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + b = \dfrac{6,72}{22,4} = 0,3(2)\)

Từ (1)(2) suy ra: a = 0,1 ; b = 0,2

Vậy :

\(\%m_{Fe} = \dfrac{0,1.56}{16,8}.100\% = 33,33\%\\ \%m_{Mg} = \dfrac{0,2.24}{16,8}.100\% = 28,57\%\\ \%m_{Cu} = 100\% - 33,33\% - 28,57\% = 38,1\%\)

Chất không tan là Ag.

=> mAg= 6,25(g)

nH2=0,25(mol)

PTHH: Zn + H2SO4 -> ZnSO4 + H2

-> nZn=nH2= 0,25(mol)

=>mZn= 0,25 . 65=16,25(g)

=> 

             2Ag+ 2H₂SO₄→ Ag₂SO₄+ 2H₂O+ SO₂

(mol)  0,04                              0,02                                                       Zn+ H₂SO₄→ ZnSO₄+ H₂

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

6,25g một chất rắn không tan là: Ag₂SO₄

→ \(n_{Ag_2SO_4}=\dfrac{m}{M}=\dfrac{6,25}{312}=0,02\left(mol\right)\)

=> m_Ag=n.M=0,04.108=4,32(g)

      m_Zn=n.M= 0,25.65=16,25(g)

→ m_{hh kim loại}=m_Ag+m_Zn=4,32+16,25=20,57(g)

=> %m_Ag=\(\dfrac{4,32}{20,57}.100\%=21\%\)

     %m_Zn= 100%- 21%= 79%

         2Ag+2H₂SO₄→Ag₂SO₄+2H₂O+SO₂

(mol) 0,04                   0,02                                                          

           Zn+H₂SO₄→ZnSO₄+H₂

(mol) 0,25                              0,25                                               

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

6,25g một chất rắn không tan đó là: Ag₂SO₄

→ \(n_{Ag_2SO_4}=\dfrac{m}{M}=\dfrac{6,25}{312}=0,02\left(mol\right)\)

m_Ag=n.M=0,04.108=4,32(g)

m_Zn=n.M=0,25.65=16,25(g)

m_{hh kim loại}= m_Ag+ m_Zn=4,32+16,25= 20,57(g)

=> %m_Ag=\(\dfrac{4,32}{20,57}.100\%=21\%\)

     %m_Zn= 100%-21%= 79%