\(PTHH:2A+3H_2SO_4\rightarrow A_2\left(SO_4\right)_3+3H_2O\\ \text{Ta có }n_A=\dfrac{8,1}{M_A}\left(mol\right);n_{A_2SO_3}=\dfrac{51,3}{2M_A+288}\left(mol\right)\\ \text{Theo PT: }n_A=2n_{A_2SO_3}\\ \Rightarrow\dfrac{8,1}{M_A}=\dfrac{102,6}{2M_A+288}\\ \Rightarrow16,2M_A+2332,8=102,6M_A\\ \Rightarrow86,4M_A=2332,8\\ \Rightarrow M_A=27\)
Vậy A là nhôm (Al)