Excess and Limiting Reactants

    A balanced equation tells us what should happen in a reaction and how much of something is produced. However, sometimes it is not possible for every atom/molecule of the reactants to combine so it is needed to add more of one reactant than the other.

    Just think of a recipe. You need 3 eggs and 1 cup of cheese to make a cheese omelet. You have 6 eggs and 1 cup of cheese. Therefore, you can only make 1 omelet and you'll still have 3 eggs left over but no cheese. That would mean the egg is the Excess Quantity and the cheese is the Limiting Reactant.

  

Now lets learn how to calculate the amount of product produced when there's an excess reactant.

e.g. How many grams of OCl2 will be formed when 44.0g of O2 reacts with 97.0g of Cl2.

General Process

Convert both reactants to the desired product and the smaller amount of product is the real amount of product that will be produced.

Step 1: Write a balanced equation.

O2 + 2Cl2 -> 2OCl2

Step 2: Convert Cl2 to OCl2

97.0g Cl2 x (1 mole Cl2/ 71.0g Cl2) x (2 mole OCl2/ 2 mole Cl2) x (87.0g OCl2/ 1 mole OCl2) = 

118.6g OCl2.

119g of OCl2 would be produced if 97g of Cl2 reacted with sufficient O2.

Step 3: Convert O2 to OCl2

44.0g O2 x (1 mole O2/ 32.0g O2) x (2 moles OCl2/ 1 mole O2) x (87.0g OCl2/ 1 mole OCl2) = 

239.25g OCl2.

239g of OCl2 would be produced if 44.0g of O2 reacted with sufficient Cl2.

Since 97g of Cl2 reacts to produce the smaller amount of OCl2 product, the Cl2 is the Limiting Reactant and the O2 is the Excess Reactant, and 119g OCl2 would be expected to be produced.

Finding the amount of excess is very similar with one added step.

e.g When 41g of O2 is reacted with 164g of Cl2 how many grams of which reactant will be excess?

General Process

Convert one of the reactants to the other reactant to see which is excess and which is limiting. Determine which is left over and by how much.

Step 1: Write a balanced equation.

O2 + 2Cl2 -> 2OCl2

Step 2: TO find which reactant is in excess calculate how many grams of chlorine gas would be required to react with 41g of oxygen gas.

41.0g x (1 mole O2/ 32g) x (2 mole Cl2/ 1 mole O2) x (71g Cl2/ 1 mole Cl2) = 181.93g Cl2

182g of Cl2 gas would be required to react with 41.0g of O2 gas.

Chlorine is the limiting quantity and Oxygen is excess quantity.

However, to calculate how much is excess we must find out how much of the second reactant (O2 gas) would react with the limiting reactant (164g Cl2)

Step 3: Convert Cl2 to O2 to see how much O2 would be needed to react with 164g of Cl2.

164g Cl2 x (1 mole Cl2/ 71g Cl2) x (1 mole O2/ 2 mole Cl2) x (32g O2/ 1 mole O2) = 36.958g O2

37g of O2 would react with 164g of Cl2.

 41.0g O2 (have)

-37.0g O2 (reacts)

4.0g O2 (in excess)

Solution: 4.0g O2 gas would be in excess when 164g Cl2 gas reacts with 41.0g O2 gas.

 Thanks for reading!