PTHH: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2 (1)
CuO + H2 ---to---> Cu + H2O (2)
Ta có: \(n_{CuO}=\dfrac{4,8}{80}=0,06\left(mol\right)\)
Theo PT(2): \(n_{H_2}=n_{CuO}=0,06\left(mol\right)\)
Theo PT(1): \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,06=0,04\left(mol\right)\)
=> \(m_{Al}=0,04.27=1,08\left(g\right)\)