Đặt công thức oxit kim loại là MxOy a (mol)
Ta có: \(O+H_2\rightarrow H_2O\)
nO=ya=nH2(1)=0,044(mol)
moxit=mKL+mO=mKL+16.0,044=2,552(g) ⇒ mKL=1,848(g)
\(M+nHCl\rightarrow MCl_n+\dfrac{n}{2}H_2\)
nM=xa=(2/n).nH2=0,066/n(mol)
\(M_M=\dfrac{1,848}{\dfrac{0,066}{n}}=28n\Rightarrow n=2\Rightarrow M_M=56\left(g/mol\right)\Rightarrow Fe\)
Ta có: \(\dfrac{xa}{ya}=\dfrac{x}{y}=\dfrac{\dfrac{0,066}{2}}{0,044}=\dfrac{3}{4}\)
⇒Fe3O4