Consider the following reaction: 4Fe + 302 → 2Fe2O3. What mass of iron(III) oxide would be produced if 250 g of Fe were reacted with 120 g of oxygen?

Mass = 357.7 g

Explanation:

Given data:

Mass of Fe =  250 g

Mass of oxygen = 120 g

Mass of iron(III) oxide produced = ?

Solution:

Chemical equation:

4Fe + 3O₂        →     2Fe₂O₃

Number of moles of Fe:

Number of moles = mass/molar mass

Number of moles = 250 g/ 55.8 g/mol

Number of moles = 4.48 mol

Number of moles of O₂ :

Number of moles = mass/molar mass

Number of moles = 120 g/ 32 g/mol

Number of moles = 3.75 mol

Now we will compare the moles of reactants with product.

        Fe          :          Fe₂O₃

        4            :             2

      4.48         :        2/4×4.48 = 2.24

        O₂          :          Fe₂O₃

        3            :             2

      3.75         :        2/3×3.75= 2.5

Less number of moles of Fe₂O₃ are produced by Fe thus it will act as limiting reactant.

Mass of Fe₂O₃:

Mass = number of moles × molar mass

Mass = 2.24 mol × 159.69 g/mol

Mass = 357.7 g