Question

If 43.7 g of iron is completely used to produce 62.5 g of iron (III) oxide, how many grams of oxygen are involved in the reaction?

Answer 1

Mass of Oxygen : 18.8 g

Further explanation

Reaction(balanced) :

4Fe + 3O₂ → 2Fe₂O₃

mass Fe = 43.7 g

mol Fe(MW= 55,845 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{43.7}{55.845}\\\\mol=0.783$

mol O₂ : mol Fe = 3 : 4, so mol O₂ :

$\tt \dfrac{3}{4}\times 0.783=0.5873$

Mass O₂(MW=32 g/mol) :

$\tt mass=mol\times MW\\\\mass=0.5873\times 32\\\\mass=18.7936~g\approx 18.8$

Or simply you can Conservation of mass, where the masses before and after the reaction are the same

mass reactants=mass products

mass iron+mass oxygen=mass  iron (III) oxide

43.7 g + mass oxygen=62.5 g

mass oxygen = 62.5 - 43.7 = 18.8 g