Question

If you start with 227.8 grams of iron and 128 grams of oxygen to produce iron oxide, what is the limiting reagent?

Answer 1

The balanced chemical equation between iron and oxygen to produce iron (III) oxide is,

$4Fe(s) + 3O_{2}(g) ---> 2Fe_{2}O_{3}(s)$

Mass of Fe = 227.8 g

Moles of Fe = $227.8 g Fe * \frac{1 mol Fe}{55.85 g Fe} = 4.079 mol Fe$

Mass of oxygen = 128 g

Moles of $O_{2} = 128 g O_{2}*\frac{1 mol O_{2}}{32 g O_{2}}= 4mol O_{2}$

Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.

Mass of iron (III) oxide produced from Iron = $4.079 mol Fe * \frac{2 mol Fe_{2}O_{3}}{4 mol Fe} *\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 325.7 g Fe_{2}O_{3}$

Mass of iron (III) oxide produced from oxygen=$4 mol O_{2}*\frac{2 molFe_{2}O_{3}}{3 mol O_{2}}*\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 425.84 g Fe_{2}O_{3}$

Iron (Fe) produces the least amount of the product iron (III) oxide. So, Fe is the limiting reactant.