\(\%_{Al}=\dfrac{27.2}{27.2+16.3}.100\%=\dfrac{900}{17}\%\\ \Rightarrow m_{Al}=20,4.\dfrac{900}{17}\%=10,8(g)\\ \Rightarrow m_{O}=20,4-10,8=9,6(g)\)
Chọn A
\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\\ n_{Al}=2n_{Al_2O_3}=0,4\left(mol\right)\\ \Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\\ n_O=3n_{Al_2O_3}=0,6\left(mol\right)\\ \Rightarrow m_O=0,6.16=9,6\left(g\right)\)