n HCl = \(\dfrac{m}{M}=\dfrac{36,5}{36,5}=1\left(mol\right)\)

=> n Cl^- = n HCl = 1 mol

=> m Cl^- = 1*35,5 = 35,5 (g)

m Muối = m hỗn hợp + m Cl^- = 13,4 +35,5 =48,9(g)

pthh:

Mg + 2 HCl ---> MgCl₂ + H₂

2Al + 6HCl ----> 2AlCl₃ + 3H₂

Fe + HCl ---> FeCl₂ + H₂

= 22,2 + 38,325 – 1,05 = 59,475 (gam).

a)

\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ Zn + 2HCl \to ZnCl_2 + H_2\)

b)

Gọi : \(n_{H_2} = a(mol) \Rightarrow n_{HCl} = 2a\)

Bảo toàn khối lượng :

\(13,5 + 2a.36,5 = 66,75 + 2.a\\ \Rightarrow a = 0,75\\ \Rightarrow V = 0,75.22,4 = 16,8(lít)\)

a) Mg + 2 HCl -> MgCl2  + H2

2Al + 6 HCl -> 2 AlCl3 + 3 H2

Fe + 2 HCl ->  FeCl2 + H2

Zn + 2 HCl -> ZnCl2 + H2

b) mY-mX=mCl

<=> mCl= 66,75-13,5=53,25(g)

=>nCl=53,25/35,5=1,5(mol)

=> nH2= nCl/2= 1,5/2=0,75(mol)

=>V=V(H2,đktc)=0,75.22,4=16,8(l)

a. PTHH:

Fe + 2HCl \(\rightarrow\)FeCl₂ + H₂

Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

Zn + 2HCl \(\rightarrow\)ZnCl₂ + H₂

_b.

Áp dụng định luật bảo toàn khổi lượng ta có:

m_A + m_B = m_C + m_D

m = m_C + m_D - m_B

m = 42,6 + 0,7 - 25,55

m = 17,75

Vậy khối lượng của X bằng 17,75 g

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Ta có: m _{dd tăng} = m_KL - m_H2

⇒ m_H2 = 16,6 - 15,6 = 1 (g) \(\Rightarrow n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\)

Có: 27n_Al + 56n_Fe = 16,6 (1)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=0,5\left(2\right)\)

Từ (1) và (2) ⇒ n_Al = n_Fe = 0,2 (mol)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{H_2}=1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{1.36,5}{40\%}=91,25\left(g\right)\)

⇒ m _{dd sau pư} = 91,25 + 15,6 = 106,85 (g)

Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,2.133,5}{106,85}.100\%\approx24,98\%\\C\%_{FeCl_2}=\dfrac{0,2.127}{106,85}.100\%\approx23,77\%\end{matrix}\right.\)

\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{H_2}=\dfrac{16,6-15,6}{2}=0,5mol\\ n_{HCl}=1mol\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=16,6\\ 3a+2b=1\\ a=b=0,2\\ m_{Al}=0,2.27=5,4g\\ m_{Fe}=0,2.56=11,2g\\ m_{ddsau}=15,6+\dfrac{36,5}{0,4}=106,85g\\ C\%_{AlCl_3}=\dfrac{133,5.0,2}{106,85}.100\%=24,99\%\\ C\%_{FeCl_2}=\dfrac{127.0,2}{106,85}.100\%=23,77\%\)

Em đăng sang môn Hoá nha

\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)

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la 1,4 g H2

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