\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ KL:A\left(x:hoa.tri.A\right)\\ 2A+xH_2SO_4\rightarrow A_2\left(SO_4\right)_x+xH_2\\ n_{Al}=\dfrac{2.0,45}{x}=\dfrac{0,9}{x}\left(mol\right)\\ M_A=\dfrac{8,1}{\dfrac{0,9}{x}}=9x\left(\dfrac{g}{mol}\right)\)
Xét: x=1;x=2;x=3;x=8/3 => Chọn x=3 thỏa, MA=27(g/mol)
=>A là nhôm (Al=27)