\(n_{HCl}=0,65.2=1,3\left(mol\right)\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (1)

0,1<----0,3<------0,1<-----0,15

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\) (2)

 x--------->6x-------->2x

\(MgO+2HCl\rightarrow MgCl_2+H_2O\) (3)

y--------->2y-------->y

Gọi \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{MgO}=y\left(mol\right)\end{matrix}\right.\)

Theo PTHH (1), (2), (3) có:

 \(n_{HCl\left(2\right)}+n_{HCl\left(3\right)}=n_{HCl.ban.đầu}-n_{HCl\left(1\right)}\\ \Leftrightarrow6x+2y=1,3-0,3=1\left(mol\right)\left(I\right)\)

Theo PTHH (1) có: \(m_{Al}=0,1.27=2,7\left(g\right)\\ \Rightarrow m_{Al_2O_3}+m_{MgO}=102x+40y=20,9-2,7=18,2\left(g\right)\left(II\right)\) 

Từ (I),(II) có hệ phương trình:

 \(\left\{{}\begin{matrix}6x+2y=1\\102x+40y=18,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

Trong hỗn hợp ban đầu:

\(\left\{{}\begin{matrix}m_{Al}=2,7\left(g\right)\\m_{Al_2O_3}=102x=102.0,1=10,2\left(g\right)\\m_{MgO}=40y=40.0,2=8\left(g\right)\end{matrix}\right.\)

b

\(CM_{AlCl_3}=\dfrac{0,1+2x}{0,65}=\dfrac{0,1+2.0,1}{0,65}=\dfrac{6}{13}\approx0,46M\)

\(CM_{MgCl_2}=\dfrac{y}{0,65}=\dfrac{0,2}{0,65}=\dfrac{4}{13}\approx0,31M\)

Ta có:

`M_B = 30(g//mol)`

`d_{A//B} = 1,8`

`-> M_A = d_{A//B} .M_B = 30.1,8 = 54 (g//mol)`

`a)`

`n_{Zn} = (3,25)/(65) = 0,05(mol)`

`n_{HCl} = 0,2.1,75 = 0,35(mol)`

`PTHH: Zn + 2HCl -> ZnCl_2 + H_2`

Xét tỉ lệ: `0,05 < (0,35)/2 -> HCl` dư, `Zn` tan hết

Theo PT:

`n_{HCl(pư)} = 2n_{Zn} = 0,1(mol)`

`n_{H_2} =n_{ZnCl_2}= n_{Zn} = 0,05(mol)`

`-> V_1 = 0,05.22,4 = 1,12(l)`

`-> n_{HCl(dư)} = 0,35 - 0,1 = 0,25(mol)`

`n_{Fe} = (5,6)/(56) = 0,1(mol)`

`PTHH: Fe + 2HCl -> FeCl_2 + H_2`

Xét tỉ lệ: `0,1 < (0,25)/2 -> HCl` dư, `Fe` tan hết

Theo PT: 

`n_{HCl(pư)} = 2n_{Fe} = 0,2(mol)`

`n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,1(mol)`

`-> V_2 = 0,1.22,4 = 2,24(l)`

`b)`

`n_{HCl(dư)} = 0,25 - 0,2 = 0,05(mol)`

`C_{M(ZnCl_2)} = C_{M(HCl)} = (0,05)/(0,2) = 0,25M`

`C_{M(FeCl_2)} = (0,1)/(0,2) = 0,5M`

chiến thần hoá căm bách rùi:>>

`a)`

$CuO + CO \xrightarrow{t^o} Cu + CO_2(1)$
$Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2(2)$
`Fe + H_2SO_4 -> FeSO_4 + H_2(3)`

`n_{Cu} = (3,2)/(64) = 0,05(mol)`

Theo PT `(1): n_{CuO} = n_{Cu} = 0,05(mol)`

`\%m_{CuO} = (0,05.80)/(20) .100\% = 20\%`

`\%m_{Fe_2O_3} = 100\% - 20\% = 80\%`

`b)`

`m_{Fe_2O_3} = 20-4=16(g)`

`-> n_{Fe_2O_3} = (16)/(160) = 0,1(mol)`

Theo PT `(2): n_{Fe} = 2n_{Fe_2O_3} = 0,2(mol)`

Theo PT `(3): n_{H_2SO_4} = n_{Fe} = 0,2(mol)`

`-> m_{ddH_2SO_4} = (0,2.98)/(9,8\%) = 200(g)`

`c)`

Theo PT `(1),(2): n_{CO_2} = n_{CuO} + 3n_{Fe_2O_3} = 0,35(mol)`

`PTHH: Ca(OH)_2 + CO_2 -> CaCO_3 + H_2O`

Theo PT: `n_{CaCO_3} = n_{CO_2} = 0,35(mol)`

`-> m_{CaCO_3} = 0,35.80\%.100 = 28(g)`

\(n_{Al}=2a\Rightarrow n_{Mg}=a\\ 54a+24a=7,8\\ a=0,1=n_{Mg}\\ n_{Al}=0,2mol\\ m_{Mg}=24.0,1=2,4g\\ m_{Al}=7,8-2,4=5,4g\)

\(a.\\ C\%_{sau}=\dfrac{5}{100}=\dfrac{32.0,1}{32+m_{H_2O}}\\ m_{H_2O}=32\left(g\right)\\ b.\\ C_{M\left(sau\right)}=1=\dfrac{0,2.2}{0,2+V_{H_2O}}\\ V_{H_2O}=0,2\left(L\right)=200\left(mL\right)\)

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

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

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

PTHH:

\(A+2HCl\rightarrow ACl_2+H_2\) (1)

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

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

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

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

\(\Rightarrow\) HCl dư.

\(n_{H_2}=x+1,5x=0,375\left(mol\right)\)

\(\Rightarrow x=0,15\)

Mặt khác:

\(m_{hh}=m_A+m_B=Ax+Bx=x\left(A+B\right)=7,65\left(g\right)\\ \Rightarrow A+B=\dfrac{7,65}{0,15}=51g/mol\)

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

Vậy A là kim loại Mg, B là kim loại Al.

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