\(n_{HCl}=2,75.0,4=1,1\left(mol\right)\)
Ta có
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{H2}=\frac{n_{HCl}}{2}=\frac{1,1}{2}=0,55\left(mol\right)\)
\(\rightarrow V=0,55.22,4=12,32\left(l\right)\)
BTKL: m A+mHCl=m muối+mH2
\(\rightarrow25,3+1,1.36,5=m_{muoi}+0,55.2\)
\(\rightarrow m_{muoi}=64,35\left(g\right)\)