Consider the reaction:

2 H

_{2}(g) + O

_{2}(g) → 2 H

_{2}O(l)

If 20 grams of H

_{2}gas is reacted with 96 grams of O

_{2}gas,

Which reactant is the limiting reactant?

How much of the excess reactant remains?

How much H

_{2}O 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. To review, follow the strategy outlined in How to Calculate Theoretical Yield.

The mole ratios between each reactant and the product is needed to complete the calculation:

The mole ratio between H

_{2}and H

_{2}O is 1 mol H

_{2}/1 mol H

_{2}O

The mole ratio between O

_{2}and H

_{2}O is 1 mol O

_{2}/2 mol H

_{2}O

The molar masses of each reactant and product are also needed.

molar mass of H

_{2}= 2 grams

molar mass of O

_{2}= 32 grams

molar mass of H

_{2}O = 18 grams

How much H

_{2}O is formed from 20 grams H

_{2}?

grams H

_{2}O = 20 grams H

_{2}x (1 mol H

_{2}/2 g H

_{2}) x (1 mol H

_{2}O/1 mol H

_{2}) x (18 g H

_{2}O/1 mol H

_{2}O)

All the units except grams H

_{2}O cancel out, leaving

grams H

_{2}O = (20 x 1/2 x 1 x 18) grams H

_{2}O

grams H

_{2}O = 180 grams H

_{2}O

How much H

_{2}O is formed from 96 grams O

_{2}?

grams H

_{2}O = 20 grams H

_{2}x (1 mol O

_{2}/32 g O

_{2}) x (2 mol H

_{2}O/1 mol O

_{2}) x (18 g H

_{2}O/1 mol H

_{2}O)

grams H

_{2}O = (96 x 1/32 x 2 x 18) grams H

_{2}O

grams H

_{2}O = 108 grams O

_{2}O

Much more water is formed from 20 grams of H

_{2}than 96 grams of O

_{2}. Oxygen is the limiting reactant. After 108 grams of H

_{2}O forms, the reaction stops. To determine the amount of excess H

_{2}remaining, calculate how much H

_{2}is needed to produce 108 grams of H

_{2}O.

grams H

_{2}= 108 grams H

_{2}O x (1 mol H

_{2}O/18 grams H

_{2}O) x (1 mol H

_{2}/1 mol H

_{2}O) x (2 grams H

_{2}/1 mol H

_{2})

All the units except grams H

_{2}cancel out, leaving

grams H

_{2}= (108 x 1/18 x 1 x 2) grams H

_{2}

grams H

_{2}= (108 x 1/18 x 1 x 2) grams H

_{2}

grams H

_{2}= 12 grams H

_{2}

It takes 12 grams of H

_{2}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 O

_{2}.

There will be 8 grams H

_{2}remaining.

There will be 108 grams H

_{2}O 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 product limits the reaction.

For more examples, check out Limiting Reactant Example Problem and Aqueous Solution Chemical Reaction Problem.

Test your new skills on the Theoretical Yield and Limiting Reaction Test Questions.