Unit Four: Chemical Reactions and Stoichiometry

Lesson 4: Limiting and excess reactant

• Comptencies
• Starter Activities
• Main Activites

Competencies

• Define limiting and excess reactants
• Determine limiting and excess reactants of a given chemical reaction
• Show that the amount of product of a chemical reaction is dependent on the limiting reactant

Materials

Charts with different diagrams: automobile bodies, tyres, atomic and molecular models

Starter activity (5 minutes)

You can start this lesson by discussing the analogy of tyres and assembly of an automobile body. Let the student do the following activity in groups.

The automobile body manufacturer produced six car bodies while the tyre company supplied 20 tyres.

1. How many tyres are required for each car?
2. What is the ratio of the car body to the number of tyres?
3. Are all the car bodies and the tyres used up?
4. How many cars can be assembled?
5. Is it the car body or the tyres which is in excess?
6. Which one limits the number of cars that would be produced?

After discussing this analogy students surly come up with the correct answers as it is easier to understand.

Evaluation

Which part of the car limits the amount of car that will be produced? Why?

In this case make sure that the students clearly understand that the first step to such task is finding out the ratio between the parts which are used up to make a complete product. Excess and limiting are relative terms that are based on the established ratio. Students should not have the misconception that a limiting reactant is a substance which is lower in number/quantity or excess reactant is the one which is large in quantity/number.

Main activity (25 minutes)

You may give the following task to the students to do according to the procedure below. Let them explain each step analogous to the assembly of the vehicle parts.

A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  Which one is the limiting reactant and how much excess reactant remains after the reaction has stopped?

Procedure

1. Write down balanced equation for the reaction.
2. Find out the number of moles of the reactants that combine according to the balanced equation.
3. Change the given masses of the reactants to their respective number of moles.
4. Compare the ratio of the numbers of moles of the balanced equation and the moles you calculated. Are the ratios the same or different?
5. Calculate the number of moles of NO or H2O produced using the number of moles of ammonia you calculated.
6. Repeat the calculation in procedure 5 using the number of moles of oxygen you calculated.
7. Which calculation yielded less numbers of moles of the product?
8. Which result is the correct amount of the product formed? The larger or the smaller?
9. Which reactant limits formation of the product?
10.  Do all the reactants that are mixed in the reactor converted to the product? Which one is left uncombined? How many moles of the excess reactant remain uncombined?

Hopefully students can accomplish the task because it is split into small and simple activities. However, students should be supervised and assisted where ever they faced difficulty. Finally you may check their work whether it consists of the following points for each step.

1. The balanced equation for the reaction is

   4 NH3(g) + 5 O2(g)4 NO(g) + 6 H2O(g)
   This is analogous to assembling the parts of the vehicle to make a complete vehicle.

2. The ratio of number of moles of ammonia to oxygen is 4: 5. Meaning 4 moles of ammonia     reacts with 5 moles of oxygen. This is as one car body is to four tyres, meaning 1:4.
3. Number of moles of ammonia, nNH3 = actual mass of ammonia/molar mass of ammonia

   = 2g / 17g/mole
   = 0.118mole
   Number of moles of oxygen, O2 = actual mass of oxygen/molar mass of oxygen
   = 4g/32g/mole
   = 0.125mole

4. When the ratio is compared, 4:5 for the balanced equation and 0.118:0.125 for the given data, it comes to be 0.8 and 0.944. This means the reactants are not taken according to the balanced equation. We can see that unless spare parts are taken according to the proportion of parts for one complete vehicle, some are left unassembled.
5. According to the balanced equation for the reaction, to calculate the amount of water formed using ammonia:

   4 NH3(g) + 5 O2(g)4 NO(g) + 6 H2O(g)
   4 mole NH3                produces            6 mole H2O  
   0.118 mole NH3      produces              x mole H2O
   X = 0.118 mole NH3    x  6 mole H2O/ 4 mole NH3 
   = 0.177 mole H2O

6. The amount of water formed using oxygen:

   4 NH3(g) + 5 O2(g)4 NO(g) + 6 H2O(g)
   5 mole O2 produces 6mole H2O
   0.125 mole O2     produces    y
   Y = 0.125 mole O2         x 6mole H2O/5 mole O2
   = 0.15 mole H2O

1. Calculation based on the amount of oxygen produces less amount of the product, water.
2. The smaller amount of product is the correct yield of the reaction because the larger amount  is the result of the excess reactant
3. The amount of oxygen limits formation of the product. This means there is no enough amount of oxygen to combine with all the ammonia available in the reaction.
4. All the reactants are not converted to the product. Some amount of the excess reactant, ammonia, is left over in the reaction mixture uncombined. There are two different ways of calculating the amount of ammonia left uncombined. One is using the amount of product (water or nitric oxide) formed in the reaction while the second is using the amount of the limiting reactant, oxygen.

Let us use the amount of oxygen. Since we know that all the oxygen taken in the reaction combines to form the products, the amount of ammonia combined with the 0.125 moles of oxygen can be calculated as:

4 NH3(g) + 5 O2(g)

4 NO(g) + 6 H2O(g)
4 moles NH3   reacts with   5 moles O2
X   reacts with 0.125 mole O2

Mole of NH3, x

= 4 moles NH3 x 0.125 moles O2/5 moles O2
= 0.1 mole NH3

This shows only 0.1 moles ammonia reacted with 0.125 mole of oxygen. Therefore, the amount of NH3 is:

0.118 mole – 0.1 mole = 0.018mole NH3 or
17g/mole x 0.018mole = 0.306g NH3

Evaluation

Students can be asked the following questions to check their understanding.

• What are limiting and excess reactants?
• Why does the amount of the product formed depend on the limiting reactant?

Concluding activity

Let the students discuss

• The importance of balanced equation in the calculation of amounts of reactants and products and
• What happens if the reactants are not mixed according to the proportion established by the balanced equation?

Students should come to the conclusion that unless the equation for a chemical reaction is balanced, one cannot determine the proportion of the reactants that are combined. Balanced equation is a pillar for stoichiometric relationships. If reacting substances are not combined according to the ratio of the balanced equation, one of the reactant becomes excess and the other limited.