gọi CTTQ của oxit sắt cần tìm là Fe2Ox
có: nCaCO3= \(\frac{20}{100}\)= 0,2( mol)
PTPU
Fe2Ox+ xCO\(\xrightarrow[]{to}\) 2Fe+ xCO2
.\(\frac{0,2}{x}\)..................................0,2.... mol
Ca(OH)2+ CO2\(\rightarrow\) CaCO3\(\downarrow\)+ H2O
...................0,2..........0,2................. mol
\(\Rightarrow\) MFe2Ox= \(\frac{11,6}{\frac{0,2}{x}}\)= 58x( g/mol)
\(\Rightarrow\) 56. 2+ 16x= 58x
\(\Rightarrow\) x= \(\frac{8}{3}\)
\(\Rightarrow\) CTHH: Fe3O4