\(Đặt.CTTQ.kim.loại:R\\ R+2H_2O\rightarrow R\left(OH\right)_2+H_2\\ n_{H_2}=\dfrac{0,168}{22,4}=0,0075\left(mol\right)\\ n_R=n_{H_2}=0,0075\left(mol\right)\\ M_R=\dfrac{0,3}{0,0075}=40\left(\dfrac{g}{mol}\right)\\ \Rightarrow R\left(II\right):Canxi\left(Ca=40\right)\)