1. Notes Ch. 12.3 Limiting Reagent and Percent Yield
Chemistry

2. Limiting and Excess Reagents
When you are making a recipe, like lasagna, you may have more than enough of all of the ingredients but only have half of a box of lasagna noodles then you are limited by the amount of noodles as to how much lasagna you can make. Noodles would be the limiting ingredient.
Chemists also have the same situation. In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.
A balanced chemical equation is like a recipe.

3. Remember that the coefficients of a balanced chemical equation will give you the ratio of representative particles and the mole ratio.
In the reaction for ammonia: N2 + 3H2  2NH3
It tells us when I molecule (mole) of nitrogen reacts with 3 molecules (moles) of hydrogen it will produce 2 molecules (moles) of ammonia.
What would happen if two molecules (moles) of nitrogen reacted with 3 molecules (moles) hydrogen? Would you get more ammonia?
According to the equation you have to have a ratio of nitrogen to hydrogen of 1:3. The above idea gives us a ratio of 2:3. What would happen is we would have unused nitrogen along with 2 molecules (moles) of ammonia.

4. In the reaction before only hydrogen is completely used up so it is the limiting reagent, or it determines the amount of product that can be formed by a reaction. The reaction will occur only until the limiting reagent is used up.
The reactant that is not completely used up in the reaction is called the excess reagent. In the example it is nitrogen.
Sometimes we are given quantities other than moles. If we want to determine which is the limiting reagent we first must convert each reactant to moles.

5. Example
Copper reacts with sulfur to form copper(I) sulfide according to the following balanced chemical equation.
2Cu + S  Cu2
What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?
We have to convert both products into moles.
80.0 g Cu x 1 mol Cu/63.5 g Cu = 1.26 mol Cu
25.0 g S x 1 mole S/ 32.1 g S = mol S
We then convert the amount of copper to sulfur using the mole ratio. That will tell us how much copper we need.
1.26 mol Cu x 1 mol S/2 mol Cu = mol S. We have more sulfur that this so sulfur is the excess reagent and copper is the limiting reagent.

6. We can also use the limited reagent to figure out how much product we are going to make.
What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0 g S?
We know the limiting reagent from the example before. We will use the limiting reagent to calculate the amount of Cu2S formed.
1.26 mol Cu x 1 mol Cu2S/2 mol Cu x g Cu2S/1 mol Cu2S = 1.00 x 102 g Cu2S.
We will go over this in class!!!