a/ $m_{Fe}=\dfrac{160.70}{100}=112(g)$
$→n_{Fe}=\dfrac{112}{56}=2(mol)$
$m_{O}=\dfrac{160.30}{100}=48(g)$
$→n_O=\dfrac{48}{16}=3(mol)$
Suy ra CTHH: $Fe_2O_3$
b/ $m_{Al}=\dfrac{342.15,8}{100}=54(g)$
$→n_{Al}=\dfrac{54}{27}=2(mol)$
$m_S=\dfrac{342.28,1}{100}=96(g)$
$→n_S=\dfrac{96}{32}=3(mol)$
$m_O=\dfrac{342.56,1}{100}=192(g)$
$→n_O=\dfrac{192}{16}=12(mol)$
Suy ra CTHH: $Al_2S_3O_{12}$
c/ $m_{Fe}=\dfrac{152.36,84}{100}=56(g)$
$→n_{Fe}=\dfrac{56}{56}=1(mol)$
$m_{S}=\dfrac{152.21,05}{100}=32(g)$
$→n_S=\dfrac{32}{32}=1(mol)$
$m_O=\dfrac{152.42,11}{100}=64(g)$
$→n_O=\dfrac{64}{16}=4(mol)$
Suy ra CTHH: $FeSO_4$
