\(n_{Fe}=\dfrac{36,4}{56}=0,65\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

           0,65->1,3----->0,65--->0,65

=> \(\left\{{}\begin{matrix}a,V_{ddHCl}=\dfrac{1,3}{0,5}=2,6\left(l\right)\\b,V_{H_2}=0,65.22,4=14,56\left(l\right)\end{matrix}\right.\)

c, \(C_{M\left(FeCl_2\right)}=\dfrac{0,65}{2,6}=0,25M\)

\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

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

             1          2              1          1

           0,25     0,5                        0,25

b) \(n_{H2}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,25.22,4=5,6\left(l\right)\)

c) \(n_{HCl}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)

\(V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)=250\left(ml\right)\)

 Chúc bạn học tốt

\(a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\ \Rightarrow n_{H_2}=0,25\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,25\cdot22,4=5,6\left(l\right)\\ c,n_{HCl}=2n_{Mg}=0,5\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,5}{2}=0,25\left(l\right)\)

\(a,n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

          0,125->0,25----->0,125->0,125

\(\Rightarrow\left\{{}\begin{matrix}a,V_{ddHCl}=\dfrac{0,25}{0,5}=0,5\left(l\right)\\b,V_{H_2}=0,125.22,4=2,8\left(l\right)\\c,C_{M\left(ZnCl_2\right)}=\dfrac{0,125}{0,5}=0,25M\end{matrix}\right.\)

\(a.CaO+2HCl\rightarrow CaCl_2+H_2O\\ n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{CaCl_2}=n_{CaO}=0,2\left(mol\right)\\ m_{CaCl_2}=111.0,2=22,2\left(g\right)\)

Tên muối: Canxi clorua

\(b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)

a)

$n_{Al} = \dfrac{8,1}{27} = 0,3(mol)$

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : 

$n_{H_2} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

b) $n_{HCl} = 3n_{Al} = 0,9(mol)$
$\Rightarrow m_{dd\ HCl} = \dfrac{0,9.36,5}{3,65\%} = 900(gam)$

c)

$m_{dd\ sau\ pư}= 8,1 + 900 - 0,45.2 = 907,2(gam)$

$n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,15(mol)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{907,2}.100\% = 5,65\%$

Ta có: n_HCl(trong V₁) = 0,5V₁ (mol)

n_HCl(trong V₂ ) = 3V₂ (mol)

n_HCl(sau khi trộn) = 0,1 x 2,5 = 0,25 (mol)

=> 0,5V₁ + 3V₂ = 0,25 (1)

Lại có: Thể tích dung dịch thu được là 100(ml) = 0,1 (lít)

=> V₁ + V₂ = 0,1 (2)

Từ (1), (2), ta có hệ phương trình:

\(\left\{\begin{matrix}0,5V_1+3V_2=0,25\\V_1+V_2=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}V_1=0,02\left(l\right)=20\left(ml\right)\\V_2=0,08\left(l\right)=80\left(ml\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(a.\)

\(n_{HCl}=2n_{H_2}=0.5\cdot2=1\left(mol\right)\)

\(V_{dd_{HCl}}=\dfrac{1}{2}=0.5\left(M\right)\)

\(b.\)

\(n_X=a\left(mol\right)\)

\(\Rightarrow n_Y=2a\left(mol\right),n_Z=a\left(mol\right),n_T=a\left(mol\right)\)

\(M_X=M\left(\text{g/mol}\right)\)

\(\Rightarrow M_Y=2.7M\left(\text{g/mol}\right),M_Z=\dfrac{7M}{3}\left(\text{g/mol}\right),M_T=\dfrac{347}{60}M\left(\text{g/mol}\right)\)

\(m_{hh}=aM+2a\cdot2.7M+a\cdot\dfrac{7}{3}M+a\cdot\dfrac{347}{60}M=34.7\left(g\right)\)

\(\Rightarrow aM=2.4\)

\(n_{hh}=n_{H_2}=0.5\left(mol\right)\)

\(\Rightarrow a+2a+a+a=0.5\)

\(\Rightarrow a=0.1\)

\(M=\dfrac{2.4}{0.1}=24\left(\text{g/mol}\right)\Rightarrow Mg\)

\(Y=2.7\cdot24=65\left(\text{g/mol}\right)\Rightarrow Zn\)

\(Z=\dfrac{7}{3}\cdot24=56\left(\text{g/mol}\right)\Rightarrow Fe\)

\(T=\dfrac{347}{60}\cdot24=137\left(\text{g/mol}\right)\Rightarrow Ba\)

cai M_T phai la La ( 138,8 g/mol ) chu

Câu `3:`

`n_[Mg]=[2,4]/24=0,1(mol)`

`Mg + 2HCl -> MgCl_2 + H_2 \uparrow`

`0,1`      `0,2`               `0,1`      `0,1`       `(mol)`

`a)C%_[HCl]=[0,2.36,5]/200 . 100=3,65%`

`b)m_[MgCl_2]=0,1.95=9,5(g)`

`c)V_[H_2]=0,1.22,4=2,24(l)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Câu `4:`

`n_[Zn]=[3,25]/65=0,05(mol)`

`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

`0,05`  `0,1`            `0,05`    `0,05`       `(mol)`

`a)C%_[HCl]=[0,1.36,5]/200 .100=1,825%`

`b)m_[ZnCl_2]=0,05.136=6,8(g)`

`c)V_[H_2]=0,05.22,4=1,12(l)`

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{FeCl_2} = n_{Fe} = 0,1(mol)\\ m_{FeCl_2} = 0,1.127 = 12,7(gam)\\ d) n_{HCl} = 2n_{Fe} = 0,2(mol) \Rightarrow C_{M_{HCl}} = \dfrac{0,2}{0,1} = 2M\\ e)n_{Fe_3O_4} = \dfrac{2,32}{232} = 0,01(mol)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ 4n_{Fe_3O_4} = 0,04 < n_{H_2} = 0,1 \to H_2\ dư\\ \)

\(n_{Fe} = 3n_{Fe_3O_4} = 0,03(mol)\\ m_{Fe} = 0,03.56 = 1,68(gam)\)

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