EXCESS AND LIMITING REACTANTS

Have you ever camped outside in the forest and sat around the fire munching on your smores? Chances are, you have made your own smores before! If you didn't already know what a smore is (what a sad life you've been living in), it consists of marshmallows, chocolate, and crackers. This is the "formula" for each smore you make:


2 crackers + 1 chocolate + 1 marshmallow --> 1 smore

Now let's say you have 10 crackers, 5 chocolates, and 3 marshmallows. How many smores can you make?

You can only make THREE smores. That's not enough to feed everyone around the campfire now is it?

SOLUTION:

You have 10 crackers, so 10 divided by 2 is 5. 5 smores can be created with 10 crackers.

For every one smore you make, one chocolate is needed so 5 chocolates can create 5 smores.

BUT WAIT- there are only 3 marshmallows (the most important ingredient!!) so only three smores can be made! Therefore, the marshmallow is the limiting reactant because it is completely used up and limits how many smores (the product) can be made. The other ingredients (the reactants) are in excess, so they are the excess reactants.

Now let's get started with the lesson!

In an ideal world, where everything is perfect, balanced equations determines exactly what should happen in a chemical reaction. However, we all know that reality is a pain in the butt, and that not every reactant will be used up. Usually, there will be an excess of a certain reactant and another reactant that will be completely used up to create a product.

As stated above, the limiting reactant is one that will be completely used up and the excess quantity will be the one that will have some left over.

2Na + 1Cl₂ -->2NaCl

In this example, there are 5 chlorine (Cl2) molecules and 6 Na atoms. The equation states that two Na atoms would react with one chlorine molecule. Thus, only three molecules of chlorine is needed for 6 Na atoms. The rest of the chlorine molecules are in excess and will remain unreacted.

Example: 50.0g of Na reacts with 53.5g of Cl2. Which one is in excess and by how many grams?

Step 1: Write a balanced equation.

2Na + 1Cl₂ -->2NaCl

Step 2: Convert each reactant to the product. The one who produces the smaller amount of product will be the actual amount produced.

50.0g Na x 1mol Na/23.0g Na x 2mol NaCl/2mol Na x 58.5g NaCl/1mol NaCl

= 127.17g NaCl

53.5g Cl₂  x 1mol  Cl₂/71.0g Cl₂ x 2mol NaCl/1mol Cl₂ x 58.5g NaCl/1mol NaCl

= 88.16g NaCl <---- the actual product

(Cl2 is limited reactant because it limits the product produced)

Step 3: Convert the limiting reactant to the excess reactant to find out how much of the excess would be sufficient to react with 53.5g of Cl2.

53.5g Cl₂  x 1mol  Cl₂/71.0g Cl₂ x 2mol Na/1mol Cl₂ x 23.0g Na/1mol Na

= 34.7g Na is needed in order to create the actual product

Step 4: Subtract the amount of the excess reactant that is reacted with (or needed) from the amount that it started with before the reaction has taken place to get the excess amount.

50.0g Na - 34.7g Na = 15.3g Na is in excess.

Think you got the hang of it? Here are some practice problems for you: (have fun *evil smile*)

1) How many grams of CaCl2 will be formed when 24.3g of Ca reacts wieh 32.6g of Cl2? Which is the limiting and excess, by how much?
2) Given the balnced equation: Pb + F --> PbF2. How many grams of lead(II) flouride will be produced? Which is in excess and by how much?
3) If 20g of Beryllium is reacted with 35g of Sodium Oxide...
  a) Write the balanced equation.
  b) How much beryllium oxide can be formed?
  c) What is the limiting reactant.