Đặt `n_A = n_B = a(mol)`

`n_{HCl} = (250.18,25\%)/(36,5) = 1,25(mol)`

`n_{H_2} = (8,4)/(22,4) = 0,375(mol)`

`PTHH:`

`A + 2HCl -> ACl_2 + H_2`

`2B + 6HCl -> 2BCl_3 + 3H_2`

Theo PT: `n_{HCl(pư)} = 2n_{H_2} = 0,75(mol) < 1,25(mol)`

`-> HCl` dư, kim loại ten hết

Theo PT: `n_{H_2} = n_A + 3/2 n_B`

`-> a + 1,5a = 0,375`

`<=> a = 0,15(mol)`

`-> 0,15M_A + 0,15M_B = 7,65(g)`

`<=> M_A + M_B = 51(g//mol)`

`-> M_B < 51(g//mol)`

Mà B có hóa trị III

`-> B:Al`

`-> M_A = 51 - 27 = 24(g//mol)`

Vậy A là Mg còn B là Al

\(n_{HCl}=\dfrac{250.18,25\%}{100\%}:36,5=1,25\left(mol\right)\)

\(n_{H_2}=\dfrac{8,4}{22,4}=0,375\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_A=x\left(mol\right)\\n_B=x\left(mol\right)\end{matrix}\right.\) (theo đề)

\(A+2HCl\rightarrow ACl_2+H_2\)

x---->2x---------------->x

\(2B+6HCl\rightarrow2BCl_3+3H_2\)

x----->3x------------------->1,5x

Theo PTHH có: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.1,25=0,625\left(mol\right)>n_{H_2.thoát.ra.theo.đề}\)

\(\Rightarrow HCl.dư\\ \Rightarrow n_{H_2}=x+1,5x=0,375\\ \Rightarrow x=0,15\)

Có: \(m_{hh.kim.loại}=m_A+m_B=Ax+Bx=x\left(A+B\right)=7,65\) (g)

\(\Rightarrow A+B=\dfrac{7,65}{0,15}=51\left(g/mol\right)\)

Mà A là kim loại hóa trị II, B là kim loại hóa trị III.

\(\Rightarrow\left\{{}\begin{matrix}A=24\left(Mg\right)\\B=27\left(Al\right)\end{matrix}\right.\)

a. PTHH: R + H₂SO₄ ---> RSO₄ + H₂ (1)

Ta có: \(n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

Theo PT₍₁₎: \(n_R=n_{H_2}=0,16\left(mol\right)\)

=> \(M_R=\dfrac{3,84}{0,16}=24\left(g\right)\)

Vậy R là magie (Mg)

b. PTHH:

Mg + HCl ---> MgCl₂ + H₂ (2)

Theo PT₍₂₎: \(n_{H_2}=n_{Mg}=0,16\left(mol\right)\)

=> \(V_{H_2}=0,16.22,4=3,584\left(lít\right)\)

(Do câu b đề ko rõ lắm nên mik làm như vậy, nếu sai bn bình luận nhé.)

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

\(a.R+2H_2O\rightarrow R\left(OH\right)_2+H_2\\ b.n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\\ n_R=n_{H_2}=0,5mol\\ M_R=\dfrac{20}{0,5}=40g/mol\)
Vậy kim loại R là Ca

\(n_{H2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

         1         2              1           1

       0,03   0,06         0,03       0,03

       \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

          1            2             1              1

         0,03      0,06        0,03

a) \(n_{Mg}=\dfrac{0,03.1}{1}=0,03\left(mol\right)\)

\(m_{Mg}=0,03.24=0,72\left(g\right)\)

\(m_{MgO}=1,92-0,72=1,2\left(g\right)\)

b) Có : \(m_{MgO}=1,2\left(g\right)\)

\(n_{MgO}=\dfrac{1,12}{40}=0,03\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,06+0,06=0,12\left(mol\right)\)

400ml = 0,4l

\(C_{M_{ddHCl}}=\dfrac{0,12}{0,4}=0,3\left(l\right)\)

c) \(n_{MgCl2\left(tổng\right)}=0,03+0,03=0,06\left(mol\right)\)

\(C_{M_{MgCl2}}=\dfrac{0,06}{0,4}=0,15\left(M\right)\)

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