Limiting Reagents
Consider the burning (oxidizing) of octane, C8H18, in the combustion chamber in our
cars:
2 C8H18(g) + 25O2(g)  16 CO2(g) + 18 H2O(g)
There is an infinite source of oxygen from the air but the amount of C8H18 is limited by
how much we have in the gas tank.
Q: What will happen to the engine when the gas runs out?
The reason the engine will stop is because the above reaction stops – no more CO2 or H20
is produced.
Of the two reactants, the octane limits how far the reaction will go. It is the limiting
reagent. To make sure that all of the octane is used up there must be an excess amount of
oxygen (there will be some O2 left over after the reaction has stopped).
In most chemical reactions, unless carefully planned, there is an excess reagent and a
limiting reagent. If the word excess is not included, how can you tell which reactant is
the limiting reagent?
Determining the Limiting Reagent
1. Calculate the moles of each reactant
2. Use molar ratio to determine the limiting reagent.
Example: If 4.44 g of calcium oxide is mixed with 7.77 g of water, how many grams of
calcium hydroxide form?
4.44g
1 CaO
+
Moles of CaO = (4.44g)
= 0.0791 mol
7.77g
1 H2O 
?g
1Ca(OH)2
÷ (56.08g/mol)
÷
n H2O= (7.77g)
(18.01g/mol)
= 0.431 mol
3. Since the ratio of CaO : H2O is 1:1 we can compare the moles directly. CaO is
the limiting reagent because there is less of it.
Note: because the ratio of CaO : Ca(OH)2 is also 1:1 only 0.0791 moles of Ca(OH)2 can
be made.
Mass of Ca(OH)2 = (0.0791mol) x (74.08g/mol) = 5.86
Example:
If 79.1 g of zinc metal is added to 96.6 g of hydrochloric acid, then zinc chloride and
hydrogen gas are formed. What mass (in g) of zinc chloride would be produced?
79.1g
1 Zn(s)
Moles of Zn
96.6g
+ 2 HCl(aq)
?g
 1 ZnCl2(aq) + 1 H2(g)
÷
= (79.1g)
(65.39g/mol)
= 1.21 mol
÷
Moles of HCl = (96.6g)
(36.45g/mol)
= 2.65 mol
It’s tempting to say Zn is the limiting reagent because 1.21 < 2.65 BUT we must use the
molar ratio to determine what is the limiting reagent.
Zn: Hcl
1: 2
If we had 1.21 mol of zinc we would need how many HCl
1: 2
1.21 : x
Solve for x
X = 2.42. We have MORE than the 2.42 moles so therefore the ZINC is the limiting
reagent.
Since the ratio of Zn:ZnCl2 is 1:1
Mass of ZnCl2 formed is 1.21 mol x 136.29g/mol = 165g