# How to Calculate Limiting Reactant of a Chemical Reaction

## Determining the Limiting Reactant

Chemical reactions rarely occur when exactly the right amount of reactants will react together to form products. One reactant will be used up before another runs out. This reactant is known as the limiting reactant.

## Strategy

This is a strategy to follow when determining which reactant is the limiting reactant.
Consider the reaction:
2 H2(g) + O2(g) → 2 H2O(l)
If 20 grams of H2 gas is reacted with 96 grams of O2 gas,

• Which reactant is the limiting reactant?
• How much of the excess reactant remains?
• How much H2O is produced?

To determine which reactant is the limiting reactant, first determine how much product would be formed by each reactant if all the reactant was consumed. The reactant that forms the least amount of product will be the limiting reactant.

Calculate the yield of each reactant.

The mole ratios between each reactant and the product are needed to complete the calculation:
The mole ratio between H2 and H2O is 1 mol H2/1 mol H2O
The mole ratio between O2 and H2O is 1 mol O2/2 mol H2O
The molar masses of each reactant and product are also needed:
molar mass of H2 = 2 grams
molar mass of O2 = 32 grams
molar mass of H2O = 18 grams
How much H2O is formed from 20 grams H2?
grams H2O = 20 grams H2 x (1 mol H2/2 g H2) x (1 mol H2O/1 mol H2) x (18 g H2O/1 mol H2O)
All the units except grams H2O cancel out, leaving
grams H2O = (20 x 1/2 x 1 x 18) grams H2O
grams H2O = 180 grams H2O
How much H2O is formed from 96 grams O2?
grams H2O = 20 grams H2 x (1 mol O2/32 g O2) x (2 mol H2O/1 mol O2) x (18 g H2O/1 mol H2O)
grams H2O = (96 x 1/32 x 2 x 18) grams H2O
grams H2O = 108 grams O2O

Much more water is formed from 20 grams of H2 than 96 grams of O2. Oxygen is the limiting reactant. After 108 grams of H2O forms, the reaction stops. To determine the amount of excess H2 remaining, calculate how much H2 is needed to produce 108 grams of H2O.
grams H2 = 108 grams H2O x (1 mol H2O/18 grams H2O) x (1 mol H2/1 mol H2O) x (2 grams H2/1 mol H2)
All the units except grams H2 cancel out, leaving
grams H2 = (108 x 1/18 x 1 x 2) grams H2
grams H2 = (108 x 1/18 x 1 x 2) grams H2
grams H2 = 12 grams H2
It takes 12 grams of H2 to complete the reaction. The amount remaining is
grams remaining = total grams - grams used
grams remaining = 20 grams - 12 grams
grams remaining = 8 grams
There will be 8 grams of excess H2 gas at the end of the reaction.
There is enough information to answer the question.
The limiting reactant was O2.
There will be 8 grams H2 remaining.
There will be 108 grams H2O formed by the reaction.

Finding the limiting reactant is a relatively simple exercise. Calculate the yield of each reactant as if it were completely consumed. The reactant that produces the least amount of product limit the reaction.

## More

For more examples, check out Limiting Reactant Example Problem and Aqueous Solution Chemical Reaction Problem. Test your new skills by answering Theoretical Yield and Limiting Reaction Test Questions.

## Sources

• Vogel, A. I.; Tatchell, A. R.; Furnis, B. S.; Hannaford, A. J.; Smith, P. W. G. Vogel's Textbook of Practical Organic Chemistry, 5th Edition. Pearson, 1996, Essex, U.K.
• Whitten, K.W., Gailey, K.D. and Davis, R.E. General Chemistry, 4th Edition. Saunders College Publishing, 1992, Philadelphia.
• Zumdahl, Steven S. Chemical Principles, 4th Edition. Houghton Mifflin Company, 2005, New York.
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