\(n_{Fe}=\dfrac{1,05}{56}=0,01875\left(mol\right)\)
PTHH: FexOy + yH2 --to-> xFe + yH2O
\(\dfrac{0,01875}{x}\)<---------0,01875
=> \(M_{Fe_xO_y}=56x+16y=\dfrac{1,5}{\dfrac{0,01875}{x}}=80x\left(g/mol\right)\)
=> \(24x=16y\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\)
=> CTHH: Fe2O3 (Sắt (III) oxit)
