FexOy+yCO\(\rightarrow\)xFe+yCO2(1)
CO2+Ca(OH)2\(\rightarrow\)CaCO3+H2O(2)
- Theo PTHH (2): \(n_{CO_2}=n_{CaCO_3}=\dfrac{7}{100}=0,07mol\)
- Theo PTHH(1) ta thấy: Ooxit+OCO=OCO2
\(\rightarrow\)nO(oxit)=nO(CO2)-nO(CO)=0,07.2-0,07=0,07 mol
mO(oxit)=0,07.16=1,12 gam
m=mFe=4,06-1,12=2,94 gam\(\rightarrow\)nFe=\(\dfrac{2,94}{56}=0,0525mol\)
\(\dfrac{x}{y}=\dfrac{n_{Fe}}{n_O}=\dfrac{0,0525}{0,07}=\dfrac{3}{4}\)
Fe3O4