\(\text{Gọi} \ n_{Al} = 2a(mol) \Rightarrow n_{Zn} = a(mol)\\
\text{Ta có : }\ 2a.27 + 65a = 1,19 \Rightarrow a = 0,01(mol)\\
\text{Bảo toàn electron : } 3n_{Al} + 2n_{Zn} = 2n_{O(oxit)} \Rightarrow n_O = \dfrac{0,02.3 + 0,01.2}{2} = 0,04(mol)\\
n_{H^+} = n_{HCl} = 0,071.1 = 0,071(mol)\\
2H^+ + O_{oxit} \to H_2O\\
n_{O\ pư} = \dfrac{1}{2}n_{H^+} = 0,0355(mol)\\
\Rightarrow n_{O\ dư} = 0,04 - 0,355 = 0,0045(mol)\\
\Rightarrow m_{oxit\ dư} = 1,19 + 0,0045.16 = 1,262(gam)\)