\(n_A=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ \Rightarrow M_A=\dfrac{m}{n}=\dfrac{80}{0,5}=160\left(g\right)\)
\(CTHH.của.A.có.dạng:X_2O_3\\ \Leftrightarrow X.2+16.3=160\\ \Leftrightarrow X=56\left(đvC\right)\\ \Rightarrow X.là.Fe\left(sắt\right)\)