\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có :
\(n_{H2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Theo phương trình : \(n_{HCl}=2n_{H2}=1\left(mol\right)\)
\(\rightarrow m_{HCl}=1.36,5=36,5\left(g\right)\)
mmuối = mkim loại + mHCl - mH2
\(\rightarrow m_{muoi}=20+36,5-0,5.2=55,5\left(g\right)\)