A balanced chemical equation shows the molar amounts of reactants that will react together to produce molar amounts of products. In the real world, reactants are rarely brought together with the exact amount needed. One reactant will be completely used up before the others. The reactant used up first is known as the limiting reactant. The other reactants are partially consumed where the remaining amount is considered "in excess". This example problem demonstrates a method to determine the limiting reactant of a chemical reaction.
Sodium hydroxide (NaOH) reacts with phosphoric acid (H3PO4) to form sodium phosphate (Na3PO4) and water (H2O) by the reaction:
3 NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l)
If 35.60 grams of NaOH is reacted with 30.80 grams of H3PO4,
a. How many grams of Na3PO4 are formed?
b. What is the limiting reactant?
c. How many grams of the excess reactant remains when the reaction is complete?
Molar mass of NaOH = 40.00 grams
Molar mass of H3PO4 = 98.00 grams
Molar mass of Na3PO4 = 163.94 grams
To determine the limiting reactant, calculate the amount of product formed by each reactant. The reactant the produces the least amount of product is the limiting reactant.
To determine the number of grams of Na3PO4 formed:
grams Na3PO4 = (grams reactant) x (mole of reactant/molar mass of reactant) x (mole ratio: product/reactant) x (molar mass of product/mole product)
Amount of Na3PO4 formed from 35.60 grams of NaOH
grams Na3PO4 = (35.60 g NaOH) x (1 mol NaOH/40.00 g NaOH) x (1 mol Na3PO4/3 mol NaOH) x (163.94 g Na3PO4/1 mol Na3PO4)
grams of Na3PO4 = 48.64 grams
Amount of Na3PO4 formed from 30.80 grams of H3PO4
grams Na3PO4 = (30.80 g H3PO4) x (1 mol H3PO4/98.00 grams H3PO4) x (1 mol Na3PO4/1 mol H3PO4) x (163.94 g Na3PO4/1 mol Na3PO4)
grams Na3PO4 = 51.52 grams
The sodium hydroxide formed less product than the phosphoric acid. This means the sodium hydroxide was the limiting reactant and 48.64 grams of sodium phosphate is formed.
To determine the amount of excess reactant remaining, the amount used is needed.
grams of reactant used = (grams of product formed) x (1 mol of product/molar mass of product) x (mole ratio of reactant/product) x (molar mass of reactant)
grams of H3PO4 used = (48.64 grams Na3PO4) x (1 mol Na3PO4/163.94 g Na3PO4) x (1 mol H3PO4/1 mol Na3PO4) x (98 g H3PO4/1 mol)
grams of H3PO4 used = 29.08 grams
This number can be used to determine the remaining amount of excess reactant.
Grams H3PO4 remaining = initial grams H3PO4 - grams H3PO4 used
grams H3PO4 remaining = 30.80 grams - 29.08 grams
grams H3PO4 remaining = 1.72 grams
When 35.60 grams of NaOH is reacted with 30.80 grams of H3PO4,
a. 48.64 grams of Na3PO4 are formed.
b. NaOH was the limiting reactant.
c. 1.72 grams of H3PO4 remain at completion.
For more practice with limiting reactants, try the Limiting Reactant Printable Worksheet (pdf format).
Worksheet answers (pdf format)
Also try the Theoretical Yield and Limiting Reactant test. Answers appear after the final question.