a. PTHH: \(4CO+Fe_3O_4\rightarrow3Fe+4CO_2\)
b. \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(\Rightarrow n_{Fe_3O_4}=0,1\left(mol\right)\) ⇒ \(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
c. \(n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\)
⇒ \(n_{CaCO_3}=0,4\left(mol\right)\Rightarrow m_{CaCO_3}=100.0,4=40\left(g\right)\)