\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH:
`Mg + 2HCl -> MgCl_2 + H_2`
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
`MgCl_2 + 2NaOH -> Mg(OH)_2 + 2NaCl`
`AlCl_3 + 3NaOH -> Al(OH)_3 + 3NaCl`
Theo PTHH: `n_{NaOH} = n_{-Cl} = 2n_{H_2} = 0,4 (mol)`
`=> V = (0,4)/(0,25) = 1,6M`