\(n_{H_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)
PTHH:
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
`Mg + 2HCl -> MgCl_2 + H_2`
`Fe + 2HCl -> FeCl_2 + H_2`
Theo PT: `n_{HCl} = 2n_{H_2} = 1,6 (mol)`
Theo ĐLBTKL:
`m_{KL} + m_{HCl} = m_{muối} + m_{H_2}`
`=> m_{muối} = 25,4 + 1,6.36,5 - 1,6.2 = 80,6 (g)`