\(n_{CO2}=n_{CaCO3}=\frac{22,5}{100}=0,225\left(mol\right)\)
Ta có:
\(n_{O.trong.oxit}=n_{CO2}=0,225\left(mol\right)\)
\(\Rightarrow m_{Fe}=12-0,225.16=8,4\left(g\right)\)
\(n_{Fe}=\frac{8,4}{56}=0,15\left(mol\right)\)
Ta có
\(n_{Fe}:n_O=0,5:0,225=2:3\)
Vậy CTHH là Fe2O3