CHEMICAL REACTIONS

8.5 - Calculations based on Chemical Equations

8.5.4 - Limiting Reactant Calculations

Often in many chemical processes a reactant is present in excess. The reactant that is present in excess is the chemical that remains after the reaction is completed. The reactant that is fully consumed during the reaction is called the limiting reagent. In combustion reactions, oxygen gas is usually present in excess in order for the the reaction to under complete combustion. Complete combustion means that the products of the reaction are CO₂ (g) and H₂O (l). Contrast to complete combustion is incomplete combustion. Incomplete combustion tends to form CO (g) as well as CO₂ (g).

In the previous sections, we took a detailed look at how to carry out Mole-mole, Mass-mass Mass-mole type calculations.

Using the same example from section 8.5.1, we now pose another question. This time you will be given the mass of both reactants (C₂H₅OH and O₂) and you will be asked to identify the limiting reagent and calculate the moles of the product (carbon dioxide gas) formed.

(a) How many grams of carbon dioxide gas is produced when 20.5 grams of ethyl alcohol, C₂H₅OH, burns with 100.0 grams of oxygen, O₂?
(b) Identify the limiting reagent and how much of the other reactant is left over?

The steps involved to answer this question is modified from the example in section 8.5.3:
(Highlighted in red are the modifications needed in order to answer this question.)

Step 1: Identify the chemical equation involved: a combustion reaction.
Step 2: Write the balanced chemical equation (review balancing equations, section 8.3).
Step 2a: Convert the mass of C₂H₅OH to moles of C₂H₅OH.
Step 2b: Convert the mass of O₂ to moles of O₂.
Step 3: Determine the stoichiometric ratio of the substances that you are working with, specifically:

• between C₂H₅OH and O₂
• between C₂H₅OH and CO₂
• between O₂ and CO₂.
  

Step 4: Using the moles of C₂H₅OH and O₂ calculated in Step 2a and 2b, compare the mole ratio of C₂H₅OH : O₂ to that obtained in Step 3. Determine which is present as the limiting reagent and which is present in excess.
Step 5: Calculate the amount of reactant that is left over.

Let's analyze:

1. Steps 1, 2, and 2a are the same as in sections 8.5.1, 8.5.2, and 8.5.3.
2. We add Step 2b because we need to convert the given mass of O₂, 100.0 g, to moles of O₂.
3. We modify Step 3 because the question is requires the knowledge of the stoichiometric ratio of C₂H₅OH : O₂ , and potentially, C₂H₅OH : CO₂ and O₂ : CO₂.
4. In Step 4, we compare the stoichiometric ratio obtained from Step 2a and 2b and determine the limiting reagent. After that is determined, we use the available number of moles of the limiting reagent to calculate the number of moles of CO₂ produced.
5. In Step 5, we calculate the amount of the reactant that is left over. Because it didn't specify the units that we need to report this quantity, we will leave the amount of reactant left over in moles.

Summary:

Refer to section 8.5.2 and section 8.5.3 for a detailed look at Step 1, 2, and 2a. Click for a detailed look at:

• Step 1 to 3,
• Step 4, and
• Step 5 of my work for this question.
  

Mass of C2H5OH (20.5 grams)	step 2a	moles of C2H5OH	step 4	limiting reagent = ?	step 4	grams of CO2 produced
Mass of O2 (100.0 grams)	step 2b	Moles of O2		reactant in excess = ?	step 5	moles of excess reactant left

(Note: We use the calculator answer in our calculations up until the final answer when follow significant figure rules to remove the excess digits.)

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Section 10.4
The Limiting Reagent Examples ..p254

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