Excess & Limiting Reactants

Sometimes, in reactions, certain conditions aren't present for a reaction to occur (ex. temperature, pressure etc..)

Because it's not always possible for every atom and molecule to form together during a reaction, we have to add more of one reactant in order to match up the atoms/molecules

This leaves us with:

-one reactant, the one that's used up completely, being the LIMITING REACTANT

-the other reactant, the one in excess is called the EXCESS QUANTITY


Still confused? Use this example to help!

You are selling lemonade at a lemonade stand. You buy 20 lemons and 103 sugar cubes. Each pitcher needs 5 lemons and 22 sugar cubes to make the perfect pitcher of lemonade.

1. How many pitchers can you make?
2. What is the excess ingredient? How much is it in excess?

-Well, since it takes 5 lemons per pitcher, and you have 20 lemons, you can make a total of FOUR    PITCHERS
-Since you're making 4 pitchers, and 22 x 4 = 88, and 103-88= 15, you have a total of 15 ICE CUBES IN EXCESS

Make more sense? :) Let's apply it to some chemistry!

Example:

5.00 g of oxygen gas reacts with 3.00 g of ammonia to produce nitrogen oxide and water. Which reactant is the limiting reactant? How much is in excess?

1) First step, write a balanced equation

                                                           5O₂ + 4NH₃ --> 4NO + 6H₂O

2) Next step, convert the grams of eachs reactant into one of the products using the number of moles in the balanced equation

5.00 g O₂ x  1 mol O₂  x  4 mol NO x  30.0 g NO  = 3.75 g NO
                    32.0 g O₂     5 mol O₂      1 mol NO

3.00 g NH₃ x 1 mol NH₃ x 4 mol NO x 30.0 g NO  =  5.294 g NO
                     17.0 g NH₃   4 mol NH₃  1 mol NO

Since there's more NO produced from the NH₃ than the O₂, we can conclude that the excess quantity is NH₃ and the limiting reactant is O₂.

3) To find how much there is of the excess, take the grams of the limiting quantity and convert it to the excess quantity. This helps us find the actual amount used in the reaction

5.00 g O₂ x 1 mol O₂ x 4 mol NH₃ x 17.0 g NH₃  = 2.125 g NH₃
                   32.0 g O₂    5 mol O₂      1 mol NH₃

Then, subtract original amount from the actual amount used up in the reaction:

                                                      3.00 g      <--original amount
                                                      2.125 g    <--amount actually used
                                                      0.875 g

Therefore, 0.875 g of NH₃ is left in excess

Want some more help? http://www.youtube.com/watch?v=lj-fdEqTBRM&feature=related