Gọi \(\left\{{}\begin{matrix}n_{Mg}:x\left(mol\right)\\n_{Fe}:y\left(mol\right)\end{matrix}\right.\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
x____2x___________x______x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x___2x_________x_______x
Ta có:
\(n_{H2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}24x+56y=20\\x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,25\left(mol\right)\\y=0,25\left(mol\right)\end{matrix}\right.\)
\(n_{HCl\left(pư\right)}=2x+2y=0,25.2+0,25.2=1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{MgCl2}=0,25.95=23,75\left(g\right)\\m_{FeCl2}=0,25.127=31,75\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{muoi}=m_{MgCl2}+m_{FeCl2}=23,75+31,75=55,5\left(g\right)\)
Mg+2HCl->MgCl2+H2
x-------------------------x
Fe+2HCl;->FeCl2+H2
y------------------------y
ta có :24x+56y=20
,...........x+y=11,2\22,4
=>x= y=0,25 mol
=>mHCl= 0,5+ 0,5 =1 mol
=>m muối = 0,25.95+0,25.127=151g