## Limiting Reactant Notes

Limiting Reactants

3. Conservation of Matter and Stoichiometry

a. Students know how to describe chemical reactions by writing balanced equations.

e. Students know how to calculate the masses of reactants and products in a chemical reaction from the mass of one of the reactants or products and the relevant atomic masses.

Objectives:  Students will be introduced to the concept of the limiting reactant, and will solve problems involving the concept.

Instruction: Instruction will begin with a warm-up question to introduce the concept of a limiting reactant, followed by notes on how to solve a limiting reactant problem.

Independent practice: students will work on their own on some sample problems.

Warm-up Question:

If I have 550 grams of peanut butter and a dozen slices of bread, and it takes 50 grams of peanut butter and two slices of bread to make a peanut butter sandwich, how many peanut butter sandwiches can I make?   Explain your answer.

In the Warm-up question above, we discover that we can make only 6 peanut butter sandwiches, because we realize we would run out of bread after just 6 sandwiches. Most people can easily solve the above question without much thought. Sometimes, chemistry problems are as simple as the warm-up question and other times they require a little more thought. Right now letâ€™s evaluate step by step how you came to your answer to the warm-up.

First, in order to determine how many sandwiches you can make you have to determine what you will run out of first, peanut butter or bread. You do this by determining how many sandwiches you could make if you completely used up all of your peanut butter, and compare it to how many sandwiches you could make if you used up all of your bread.

550 grams of peanut butter    X    _____1 sandwich______ =    11 sandwiches

50 grams of peanut butter

12 pieces of bread X   ___1 sandwich____   =   6 sandwiches

From our calculations, it is obvious that we will run out of bread first and that it is the bread that limits how many sandwiches we have. In chemistry we would call the bread a limiting factor, or a limiting reactant.

In a chemical reaction, often we will run out of one reactant before we run out of the other. The reactant that we run out of first is called the limiting reactant, because the reaction stops as soon as it is used up. The other reactant is said to be in excess.

Steps to solving limiting reactant problems.

1. Make sure you have a balanced chemical equation
2. Any time you are given a value in grams, convert it to moles.
3. Calculate how many moles of the desired product you could make using each of your reactants. The reactant that would produce less of your product is your limiting reactant.
4. Sometimes you will have to go from moles of product your limiting reactant would produce to grams of product.

Note: You may not always have to do every step of this process. It all depends on what information the question gives you, and what it asks you to solve for.

Example 1: If I react 2.75 moles of hydrogen with 1.75 moles of oxygen, how many moles of water will be formed?

Step 1: We must write a balanced chemical reaction

2 H2 + O2 à 2 H2O

Step 2: If we were given grams of reactants we have to convert to moles, fortunately, we already were given moles so we can skip this step.

Step 3: We need to calculate how many moles of water we would produce from the given amounts of hydrogen and oxygen.

1.75 moles O2 x 2 mol H2O = 3.50 moles of H2O

1 mol O2

2.75 moles H2 x 2 mol H2O = 2.75 moles of H2O

2 mol H2

Since we produced less water with the amount of hydrogen gas we were given, the hydrogen gas is our limiting reactant.

Step 4: If we were asked for grams of water, we would have to convert from

moles to grams. The problem only asked us to calculate moles of water, so we are finished with the problem.

Now lets try a problem were we have to use all four steps

Example 2: Lithium reacts with Nitrogen gas to produce lithium nitride. If I have 1.28 grams of lithium and 2.10 g of nitrogen, how much lithium nitride will be produced?

Step 1: Write a balanced equation

6 Li + N2 à 2Li3N

Step 2: Convert from grams of reactants to moles of reactants

1.28 g Li x 1 mole Li = 0.184 moles Li

6.94 g Li

2.10 g x 1 mole N2    = 0.0750 moles N2

28.0 g N2

Step 3: Calculate how many moles of lithium nitride would be produced using each reactant.

0.184 moles Li x 2 molesLi3N =   0.0115 moles Li3N

6 moles Li

0.0750 moles N2   x 2 moles Li3N = 0.150 moles Li3N

1 mole N2

Lithium is our limiting reactant.

Step 4: To fully answer the problem, we now need to convert the moles of lithium nitride we produced with our limiting reactant into grams of lithium nitride.

0.0115 moles Li3N x 34.82 g Li3N = 0.401 g Li3N

1 mole Li3N

Practice Problems:

1) One mole of nitrogen gas (N2) reacts with two moles of hydrogen gas (H2) to produce ammonia (NH3). What is the limiting reactant?

2) Methane gas (CH4) is burned in the air (O2) to produce carbon dioxide (CO2) and water (H2O).

A) If there is 3.20 grams of methane reacting with 3.20 grams of oxygen gas, what is my limiting reactant?

B) How many grams of water is produced in the reaction?

3) 6.73 grams of Aluminum metal react with 18.25 grams of hydrochloric acid

according to the reaction shown below.  How many grams of Aluminum chloride will be produced?

Al + HCl à ALCl3 + H2

1) One mole of nitrogen gas (N2) reacts with two moles of hydrogen gas (H2) to produce ammonia (NH3). What is the limiting reactant?

Step 1: Write a balanced equation:   N2 + 3H2 à 2NH3

Step 2: Since we are give moles, we just need to calculate which one will run out first.

1 mole N2 x 2 NH3 = 2 mole NH3           2 mol H2 x 2 NH3 = 1.33 mol 2 NH3

1 N2                                                 3 H2

Hydrogen is the limiting reactant

2) Methane gas (CH4) is burned in the air (O2) to produce carbon dioxide (CO2) and water (H2O).

A) If there is 3.20 grams of methane reacting with 3.20 grams of oxygen gas, what is my limiting reactant?

Step 1: Write a balanced equation. CH4 + 2O2 à CO2 + 2H2O

Step 2: Convert from grams of reactants to moles of reactants.

3.20 g CH4 x 1 mol CH4 = 0.200 mol CH4

16.0 g CH4

3.20 g O2 x 1 mol O2 = 0.10 mol O2

32.0 g O2

Step 3: Calculate which chemical would run out first by dividing the number of moles of each reactant by its coefficient in the balanced equation.

0.200 mol CH4 = 0.200                    0.10 mol O2  0.0500 - limiting

1                                                       2

Oxygen is the limiting reactant

B) How many grams of water is produced in the reaction?

Step 1: Convert from moles of oxygen to moles of water

0.10 moles O2 x 1 H2O = 0.050 mol H2O

2 O2

Step 2: Convert from moles of water to grams of water

0.050 mol H2O x 18.0 g H2O = 0.90 g H2O

1 mol H2O

3) 6.73 grams of Aluminum metal react with 18.25 grams of hydrochloric acid

according to the reaction shown below.   How many grams of Aluminum chloride will be produced?

Step 1: Write a balanced equation. 2Al + 6HCl à 2AlCl3 + 3H2

Step 2: Convert from grams to moles

6.73 g Al x 1 mol Al  = 0.25 mol Al    18.25 g HCl x 1 mol HCl = 0.49 HCl

27.0 g Al                                                  26.50 g HCl

Step 3: Find the limiting reactant

0.25 mol Al  = 0.12 Al                   0.49 mol HCl = 0.082 HCl - limiting

2                                                    6

Step 4: Convert from moles of the limiting reactant to moles of product.

0.49 mol HCl x 2AlCl3 = 0.16 mol AlCl3

6 HCl

Step 5: Convert from moles of product to grams of product.

0.16 mol AlCl3 x 133.50 g AlCl3 = 21.40g AlCl3

1 mol AlCl3