nFe=\(\frac{16,8}{56}\)=0,3(mol)
1, 3Fe +2O2\(\rightarrow\) Fe3O4
0,3____________0,1
\(\rightarrow\)mFe3O4= 0,1.232=23,2(g)
2,nH2= \(\frac{2,24}{22,4}\)=0,1(mol)
Fe+2HCl\(\rightarrow\) FeCl2 +H2
0,1_______________0,1
\(\rightarrow\)mFe= 0,1.56=5,6(g)