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Limiting Reactants for AP Chemistry

By — McGraw-Hill Professional
Updated on Feb 9, 2011

Practice problems for these concepts can be found at:

In the examples above, one reactant was present in excess. One reactant was completely consumed, and some of the other reactant would be left over. The reactant that is used up first is called the limiting reactant (L.R.). This reactant really determines the amount of product being formed. How is the limiting reactant determined? You can't assume it is the reactant in the smallest amount, since the reaction stoichiometry must be considered. There are generally two ways to determine which reactant is the limiting reactant:

  1. Each reactant, in turn, is assumed to be the limiting reactant, and the amount of product that would be formed is calculated. The reactant that yields the smallest amount of product is the limiting reactant. The advantage of this method is that you get to practice your calculation skills; the disadvantage is that you have to do more calculations.
  2. The moles of reactant per coefficient of that reactant in the balanced chemical equation is calculated. The reactant that has the smallest mole-to-coefficient ratio is the limiting reactant. This is the method that many use.

Let us consider the Haber reaction once more. Suppose that 50.0 g of nitrogen and 40.0 g of hydrogen were allowed to react. Calculate the number of grams of ammonia that could be formed.

First, write the balanced chemical equation:

N2(g) + 3 H2(g) → 2 NH3(g)

Next, convert the grams of each reactant to moles:

Divide each by the coefficient in the balanced chemical equation. The smaller is the limiting reactant:

For N2: 1.7848 mol N2/1 = 1.7848 mol/coefficient limiting reactant

For H2: 19.8432 mol H2/3 = 6.6144 mol/coefficient

Finally, base the stoichiometry of the reaction on the limiting reactant:

Anytime the quantities of more than one reactant are given it is probably a L.R. problem.

Let's consider another case. To carry out the following reaction: P2O5(s)+3H2O(l) → 2H3PO4 (aq) 125 g of P2O5 and 50.0 g of H2O were supplied. How many grams of H3PO4 may be produced?

Answer:

  1. Convert to moles:
  2. Find the limiting reactant
  3. The 1 mol and the 3 mol come from the balanced chemical equation. The 0.880 is smaller, so this is the L.R.

  4. Finish using the number of moles of the L.R.

Practice problems for these concepts can be found at:

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