Ta có: \(n_{gốc\left(-Cl\right)}=n_{HCl}=0,5.2=1\left(mol\right)\)
PTHH: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\) (1)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\) (2)
Theo PTHH (1), (2): \(n_{O\left(oxit\right)}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.1=0,5\left(mol\right)\)
=> \(m_{muối.clorua}=m_{oxit}-m_{O\left(oxit\right)}+m_{gốc\left(-Cl\right)}=18,2-0,5.16+1.35,5=45,7\left(g\right)\)