a) Fe+2HCl\(\rightarrow\)FeCl2+H2
nFe=nH2=\(\frac{10,08}{22,4}\)=0,45(mol)
mFe=0,45.56=25,2(g)
b)
nHCl=2nH2=2.0,45=0,9(mol)
CMHCl=\(\frac{0,9}{0,2}\)=4,5(M)
c)
FexOy+yH2\(\rightarrow\)xFe+yH2O
nFe=\(\frac{16,8}{56}\)=0,3(mol)
nH2O=nH2=0,45
\(\rightarrow\)x:y=0,3:0,45=2:3
Vậy CTHH là Fe2O3
nFe2O3=\(\frac{nFe}{2}\)=0,15(mol)
a=0,15.160=24(g)