Limiting Reactant Example Problem

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.

Example Problem

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?

Useful information:

• Molar mass of NaOH = 40.00 grams
• Molar mass of H₃PO₄ = 98.00 grams
• Molar mass of Na₃PO₄ = 163.94 grams

Read More

How to Calculate Limiting Reactant of a Chemical Reaction

By Todd Helmenstine

Solution

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

Answer

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.