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Limiting Reactants
Section 12.3
Limiting Reactants
Rarely are the reactants in a chemical equation present in the exact
ratios specified by the balanced equation. Generally, there is too much
of one and not enough of the other. The chemical reaction will stop
once one of the reactants is used up. This leaves the remainder of the
reactant that was in excess unreacted. The chemical that you ran out
of was the limiting reactant.
Limiting Factor Notesheet
Consider the following recipe:
3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder  12 muffins
A quick check of the pantry shows that the following quantities are
available. How many muffins could be made with respect to each
ingredient?
9 cups of flour could make ___36___ muffins
4 eggs could make __24__ muffins
8 cups of sugar could make __96___ muffins
10 tsp baking powder could make __60___ muffins
Limiting Factor Note sheet
3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder  12 muffins
What is the maximum number of muffins that could be made considering
ALL ingredients? __24___
Which ingredient would you run out of first? _eggs____
This ingredient would be referred to as the __limiting_ __reactant____.
The other ingredients are said to be __in __
excess .
When you run out of an ingredient, you stop producing the _product__.
Limiting Factor Note sheet
cont’d…
Consider the following “recipe” (equation):
2 mol of Al + 3 mol of Cl2  2 mol of AlCl3
A quick check of the chemistry lab shows that
the following quantities are available. How
much product could be made from each?
(How many times could you make the
recipe?)
10 moles Al ______
12 moles Cl2 _____
What is the maximum amount of AlCl3 that could be made
considering all the ingredients? ________
We would say the aluminum is _________________ and the
chlorine is _________________.
What if you were presented with grams of each reactant?
Let’s say there are 100 g of Al and 125 g of Cl2 available.
How much product could be formed? (Remember, the
coefficients in the equation are mole numbers)
100 g of Al would yield ________ mol of product.
125 g of Cl2 would yield ________ mol of product.
How much product could be made considering both
reactants? ________
The limiting factor would be ________.
Calculating the Product
when the Reactant is Limited
In the following equation:
If 200 g of sulfur reacts with 100 g of chlorine what mass of sulfur
chloride will be produced?
200
S8
256
100
+
4 Cl2
x

4S2Cl2
280
536
Sulfur
Chlorine
200 g = x g
256 g 536 g
100 g = x g
280 g 536 g
256x = 107200
x = 418.75 g
280x = 53600
x = 191.42 g
 lowest #
Steps for Calculating
Limiting Reactants
 Balance the equation
 Find the mass of each of the compounds
(molar mass x’s coefficient)
 Set up ratios for each of the reactants
and solve
 Look for the lowest number – this is your
limiting reactant
Limiting Reactant Practice I
Balance the following formula:
N2
+
H2

NH3
1. If you had 112 grams of nitrogen, what is the maximum
number of grams of NH3 that could be produced?
2. If you had 18 grams of hydrogen, what is the maximum
number of grams of NH3 that could be produced?
3. What is the maximum number of grams of NH3 that can be
produced with 112 grams of nitrogen and 18 grams of hydrogen?
5. What is the limiting reactant?
To Determine How Much More of the
Limiting Reactant is Needed to Use Up
the Excess Reactant and Finish the
Reaction:
200
100
S8
+
4 Cl2

4 S2Cl2
256
280
200 g = __x__
256 g
280 g
256x = 56000
x = 218.75 g total needed
100 g
=
already used
118.75 g
additional needed
to complete reaction