\(a,M_X=\dfrac{16}{0,1}=160\left(\dfrac{g}{mol}\right)\\ b,M_X=M_{Y_2O_3}=2M_Y+3.16=2M_Y+48\left(\dfrac{g}{mol}\right)\\ \Rightarrow2M_Y+48=160\\ \Leftrightarrow M_Y=56\left(\dfrac{g}{mol}\right)\\ \Rightarrow Y:Sắt\left(Fe=56\right)\\ \Rightarrow X:Fe_2O_3\left(Sắt\left(III\right)oxit\right)\)