Question

Answer 1

1, Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)

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

PTHH:

Zn + 2HCl ---> ZnCl₂ + H₂ (pư thế)

a         2a            a          a

Fe + 2HCl ---> FeCl₂ + H₂ (pư thế)

b        2b               b       b

2, Hợp chất tạo thành: ZnCl₂ (kẽm clorua), FeCl₂ (sắt (II) clorua) thuộc muối trung hoà

3, Theo pthh: \(n_{HCl}=2n_{H_2}=2.0,2=0,4\left(mol\right)\)

4, \(C_{M\left(HCl\right)}=\dfrac{0,4}{0,2}=2M\)

5, Hệ pt \(\left\{{}\begin{matrix}65a+56b=12,1\\a+b=0,2\end{matrix}\right.\Leftrightarrow a=b=0,1\left(mol\right)\)

\(\rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{6,5}{12,1}=53,72\%\\\%m_{Fe}=100\%-53,72\%=46,28\%\end{matrix}\right.\)

Answer 2

1.\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

   \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(\Rightarrow\)phản ứng thế ở hai pt trên.

2.\(Muối\):

\(ZnCl_2\): kẽm clorua

\(FeCl_2\): sắt (ll) clorua

3.\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\Rightarrow n_H=2n_{H_2}=0,4mol\)

\(\Rightarrow n_{HCl}=0,4mol\)

5.\(\left\{{}\begin{matrix}Zn:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}65x+56y=12,1\\BTe:2x+2y=2n_{H_2}=2\cdot0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,1\Rightarrow m_{Zn}=6,5g\\y=0,1\end{matrix}\right.\)

\(\%m_{Zn}=\dfrac{6,5}{12,1}\cdot100\%=53,72\%\)

\(\%m_{Fe}=100\%-53,72\%=46,28\%\)