\(2Al+6HCl--.2AlCl3+3H2\)
\(Fe+2HCl-->FeCl2+H2\)
\(m_{HCl}=\frac{280.18,25}{100}=51,1\left(g\right)\)
\(n_{HCl}=\frac{51,1}{36,5}=1,4\left(mol\right)\)
\(n_{H2}=\frac{26,88}{22,4}=1,2\left(mol\right)\)
\(n_{HCl}=2n_{H2}=2,4\left(mol\right)\)
=>vô lý..Bạn xem lại đề nha