\(2Al+6HCl-->2AlCl_3+3H_2\)
\(n_{H_2}=\frac{0,448}{22,4}=0,02\left(mol\right)=>n_{HCl}=2n_{H_2}=0,02.2=0,04\left(mol\right)\)
\(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}.0,02=\frac{0,04}{3}\left(mol\right)\)
a)=> \(m_{Al}=\frac{0,04}{3}.27=0,36\left(g\right)\)
b) \(C_{M_{HCl}}=\frac{0,04}{0,2}=0,2\left(M\right)\)