\(m_{O_2}=3,6-2,8=0,8\left(g\right)\\ n_{O_2}=\dfrac{0,8}{32}=0,025\left(mol\right)\\ n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\\2Fe+yO_2\rightarrow\left(t^o\right)2Fe_xO_y\\ V\text{ì}:n_{Fe}:n_{O_2}=0,05:0,025=2:1\Rightarrow x=1;y=1\\ \Rightarrow CTPT:FeO\)
