\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{25,55}{35,6}=0,7\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,7}{3}\), ta được HCl dư.
Theo PT: \(n_{HCl\left(pư\right)}=3n_{Al}=0,6\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,7-0,6=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)