Untitled - Kelly Walsh High School

Mass, Moles, and Equations 37
Quick Tip
2. We calculate the mole-to-coefficient ratio of each reactant by dividing the
moles of that reactant by its coefficient in the balanced chemical equation.
The reactant that has the smallest mole-to-coefficient ratio is the limiting
reactant. Many of us use this method.
Let us consider the rusting reaction once more. Suppose that 50.0 g of iron and
40.0 g of oxygen react. Calculate the number of grams iron(III) oxide that could
be formed.
First, write the balanced chemical equation:
4 Fe(s) 3 O 2(g) l 2 Fe 2O 3(s)
Next, convert the grams of each reactant to moles:
1 mol Fe
(50.0 g Fe)a b 0.89531927 mol Fe (unrounded)
55.846 g
(40.0 g O 2 )a 1 mol O 2
31.998 g b 1.250078 mol O 2 (unrounded)
Divide each by the coefficient in the balanced chemical equation. The smallest
result is the limiting reactant:
For Fe: 0.89531927 mol Fe/4 0.2238298 mol/coefficient Limiting reactant
For O2: 1.250078 mol O2/3 0.4166927 mol/coefficient
Finally, base the stoichiometry of the reaction on the limiting reactant:
1 mol Fe
(50.0 g Fe)a
55.846 g ba2 mol Fe2O3 4 mol Fe
159.689 g
ba b 71.5 g Fe2O3 1 mol Fe2O3 Any time you are given the amounts (grams, moles, and so on) of more than one
reactant, you will need to calculate the limiting reactant.