LIMITING REAGENT NOTES
The first step in solving a limiting reagent problem is being able to recognize that you have a
limiting reagent problem.
Example:
A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl according to the reaction:
Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O
What is the theoretical yield of MgCl2?
Is this a limiting reagent problem? One way to find out is to write down what is known about any
component of the reaction below that component:
Notice how quantities of both reactants are known. Which one will be used up first? You can't
tell, nor should you jump to any conclusions. Just because it looks like there is less
Mg(OH)2 present does not automatically mean it will be used up before all of the HCl is
consumed. This is a limiting reagent problem.
In order to find out which reactant is the limiting reagent, you have to compare them to each
other. This comparison must be done in moles! Convert each of the grams of reactants to moles:
Again, you should not jump any conclusions about which reactant is the limiting reagent. Just
because there are fewer mols of magnesium hydroxide does not mean it is the limiting reagent.
Pick any one of the reactants and calculate how many mols of the other reactant is needed to
completely use up the reactant picked. Choosing magnesium hydroxide:
Compare the mols HCl needed to the actual mols HCl available. In this case, 1.74 mol of HCl is
needed and 1.23 mol HCl is available. So, even though it appears that there are more mols of
HCl than Mg(OH)2, the HCl is the limiting reagent. The HCl will be run out before the
magnesium hydroxide and thereby limit the amount of product formed. For this reason, use the
mols of HCl to calculate the theoretical yield of magnesium chloride:
The theoretical yield is the maximum amount of product which can be produced (in an ideal
world). In the "real" world it is difficult to produce the amount obtained for the theoretical yield.
A percent yield is often used to show how close to ideality one has obtained in a chemical
LIMITING REAGENT NOTES
synthesis. Suppose in the reaction discussed a chemist actually obtained 55.4 g of MgCl2. This is
called the actual yield and would be given to you in the problem. To calculate the percent yield:
Approach 1: Find the limiting reagent by looking at the number of moles of each reactant.
1. Determine the balanced chemical equation for the chemical reaction.
2. Convert all given information into moles (most likely, through the use of molar mass as a
conversion factor).
3. Calculate the mole ratio from the given information. Compare the calculated ratio to the
actual ratio.
4. Use the amount of limiting reactant to calculate the amount of product produced.
5. If necessary, calculate how much is left in excess of the non-limiting reagent.
Approach 2: Find the limiting reagent by calculating and comparing the amount of product
each reactant will produce.
1.
2.
3.
4.
5.
6.
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent
consumed from the total mass of excess reagent given.
Example 1: Photosynthesis
Consider respiration, one of the most common chemical reactions on earth.
C6H12O6 + 6O2 → 6CO2 + 6H2O + energy
What mass of CO2 forms in the reaction of 25 grams of glucose with 40 grams of oxygen?
SOLUTION
Every 1 mol of glucose (C6H12O6) requires 6 mols of oxygen to obtain 6 mols of carbon dioxide
and 6 mols of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles.
25 g × (1 mol/180.06 g) = 0.1388 mol C6H12O6
40 g × (1 mol/ 32 g) = 1.25 mol O2
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio
to the actual ratio.
a. If all 1.25 mols of oxygen were used, there would be 1.25×1/6 or 0.208 moles
of glucose. There is only 0.1388 mols of glucose. It the limiting reactant.
1.25 mol O2 × (1 mol C6H12O6)/ 6 mol O2 = 0.208 mol C6H12O6
LIMITING REAGENT NOTES
b. If all 0.1388 moles of glucose were used, there would be 0.1388 x 6 or 0.8328
mols of oxygen. Because there is an excess of oxygen, the glucose amount is
used to calculate the amount of the products in the reaction.
0.1388 mol C6H12O6 × 6 mol O2/ 1 mol C6H12O6 = 0.8328 mol O2
If more than 6 moles of O2 are available per mole of C6H12O6, the oxygen is in excess
and glucose is the limiting reactant. If less than 6 moles of oxygen are available per mole
of glucose, oxygen is the limiting reactant. The ratio is 6 mole oxygen per 1 mole glucose,
OR 1 mole oxygen per 1/6 mole glucose.
The mole ratio is (0.8328 mol O2)/(0.208 mol C6H12O6) giving 4.004 ratio O2/C6H12O6.
Step 4: Use the amount of limiting reactant to calculate the amount of products.
For carbon dioxide produced: 0.1388 mols glucose × 6 = 0.8328 mols CO2.
Step 5: If necessary, calculate how much is left in excess.
1.25 mol – 0.8328 mol = 0.4172 moles of oxygen left over
Example 2: Oxidation of Magnesium
Calculate the mass of magnesium oxide possible if 2.4 g Mg reacts with 10.0 g O2.
Mg + O2 → MgO
SOLUTION
Step 1: Balance equation
2Mg + O2 → 2MgO
Step 2 and Step 3: Converting mass to moles and stoichiometry
1.00 mol Mg 2.00 mol MgO 40.31 g MgO
×
×
= 3.98 g MgO
24.31 g Mg
2.00 mol Mg 1.00 mol MgO
1 mol 𝑂2 2 mol MgO 40.31 g MgO
10.0 g 𝑂2 ×
×
×
= 25.2 g MgO
32.0 g 𝑂2
1 mol 𝑂2
1 mol MgO
2.40 g Mg ×
Step 4: The reactant that produces a smaller amount of product is the limiting reagent
Mg produces less MgO than does O2 (3.98 g MgO vs. 25.2 g MgO), therefore Mg is
the limiting reagent in this reaction.
Step 5: The reactant that produces a larger amount of product is the excess reagent
O2 produces more amount of MgO than Mg (25.2g MgO vs. 3.98g MgO), therefore
O2 is the excess reagent in this reaction.
Step 6: Find the amount of remaining excess reactant by subtracting the mass of the excess
reagent consumed from the total mass of excess reagent given.
2.40 g Mg ×
1.00 mol Mg 1.00 mol 𝑂2
32.0 g 𝑂2
×
×
= 1.58 g 𝑂2
24.31 g Mg 2.00 mol Mg 1.00 mol 𝑂2
OR Mass of excess reagent calculated using the mass of the product:
3.98 g MgO ×
1.00 mol MgO
1.00 mol 𝑂2
32.0 g 𝑂2
×
×
= 1.58 g 𝑂2
40.31 g MgO 2.00 mol MgO 1.00 mol 𝑂2
Mass of total excess reagent is 10.0g – 1.58g = 8.42g O2 is in excess.
LIMITING REAGENT NOTES
Example 3: Limiting Reagent
What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2?
C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2
SOLUTION
Using Approach 1:
A. 76.4g × (1 mol/ 266.72 g) = 0.286 moles C2H3Br3
49.1g × (1 mole/ 32 g) = 1.53 moles O2
B. If all of the oxygen is used up, 1.53 × 4/11 = 0.556 moles C2H3Br3 are required. There are
only 0.286 moles of C2H3Br3 available. C2H3Br3 is the limiting reagent.
Example 4: Limiting Reagent
What is the limiting reagent if 78 grams of Na2O2 were reacted with 29.4 grams of H2O?
SOLUTION
Using Approach 2:
78 g 𝑁𝑎2 𝑂2 ×
1.00 mol 𝑁𝑎2 𝑂2 4.00 mol NaOH
40.0 g 𝑁𝑎𝑂𝐻
×
×
= 80.04 g NaOH
77.96 g 𝑁𝑎2 𝑂2 2.00 mol 𝑁𝑎2 𝑂2 1.00 mol NaOH
Using either approach gives Na2O2 as the limiting reagent.
Example 5: Excess Reagent
How much of the excess reagent remains if 24.5 grams of CoO is reacted with 2.58 grams of O2?
4CoO + O2 → 2Co2O3
SOLUTION
0.0046 moles left in excess.
Example 6: Identifying the Limiting Reagent
Will 28.7 grams of SiO2 react completely with 22.6 grams of H2F2? If not, identify the limiting
reagent.
SiO2 +2H2F2 → SiF4 + 2H2O
SOLUTION
H2F2 is the limiting reagent.