UNIT # 8 Introduction Objectives 8.18.2 8.3 Problems 1 | 2 | 3 | 4

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:

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 5 of my work for this question.
If you need to see the specific steps, click on the links below.

 Mass of C2H5OH (20.5 grams) moles of C2H5OH limiting reagent = ? grams of CO2 produced Mass of O2 (100.0 grams) Moles of O2 reactant in excess = ? moles of excess reactant left

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Section 10.4
The Limiting Reagent Examples ..p254

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