Journal Entry #16
Limiting Reactants
Background
If you are in Chef Class (or just like to cook) you know that you need to have enough ingredients to
make a recipe. The same holds true for chemical reactions. Up to this point, we have assumed
excess amounts of reactants. Now we will start to do calculations where one of the reactants is
limiting. The reactant that limits the amount of product that can be formed is called the
Limiting Reactant.
Computer Simulations Revisited
How many molecules of Ammonia can be made with N2 and H2 molecules?
What type and how many molecules are left over?
Copy the Table and the equation
# of N2
Molecules
# of H2
Molecules
N2 + 3H2
# of NH3
Molecules
made
 2NH3
# of N2
Molecules
leftover
# of H2
Molecules
leftover
Limiting
Reactant
Now let’s solve a Limiting Reactant Problem
CH4 (g) + H2O (g)  3H2 (g) + CO (g)
What mass of water is required to react exactly with 249 g of methane.
(We have already learned the following steps in mass-mass conversions.
If you are comfortable with these steps then you may go right to step 4)
Step 1: Convert the mass of CH4 to moles:
249 g. / 16.0 = (15.6 mol CH4)
Step 2: Set up the ratio (CH4  H2O) = 1 / 1
249 g. CH4 x
Step 3: Determine molar mass of water: 16.0 + 2.0 = 18.0 grams H2O
(Remember, H2O is a reactant not a product in this equation)
Step 4
1 mol CH4
x 1 mol H2O x 18.0 g. H2O
= 280 gramsH2O
16.0 g. CH4
1 mol CH4
1 mol H2O
How to Identify the Limiting Reactant and Calculate the Amount of Product Made
Let’s say we have 300 grams of water instead of 280 (and there are still 249 grams of methane)
How would we be able to determine which reactant was limiting: The water or methane?
CH4 (g) + H2O (g)  3H2 (g) + CO (g)
Step 1
Calculate the moles of reactant present
CH4 = 249.0 / 16.0 = 15.6 mol
H2O = 300.0 / 18.0 = 16.7 mol
Step 2
Determine which reactant is limiting by comparing the number of moles of the reactants
Is Methane limiting?
# of moles of H2O x mole ratio = # of moles of CH4 required
16.7 mol. of H2O x 1/1 = 16.7 mol of CH4 required
Yes, there are only 15.6 mol. of methane, therefore it is the limiting reactant.
Is Water limiting?
# of moles of CH4 x mole ratio = # of moles of H2O required
15.6 mol. of CH4 x 1/1 = 15.6 mol. of H2O required
No, there are 16.7 mol. of water, therefore it is in excess, the methane will run out before the water.
Step 3
Use the Limiting Reactant to Calculate the Mass of the Product
Now let’s calculate the amount of H2 gas (product that would be produced in the original reaction:
CH4 (g) + H2O (g)  3H2 (g) + CO (g)
Since CH4 is the limiting reactant, we must use it to calculate the amount of H2 gas produced:
15.6 mol. of CH4 x 3 / 1 (mole ratio) = 46.8 mol of H2
Next, convert the # of moles of H2 to mass in grams:
46.8 mol. H2 x 2.02 g. H2 = 94.5 g. H2
1 mol H2
We can write the above completely in Dimensional Analysis as follows:
15.6 mol CH4 x
3 mol H2 x 2.02 g. H2 = 94.5 g. H2
1 mol CH4
1 mol H2
Therefore, 280 g. of H2O reacts with 249 g. of CH4 to produce 94.5 g of H2
Assignment
Do these problems in your journal
Show your work!
1. Calculate the mass in grams of the other product CO for the same reaction:
CH4 (g) + H2O (g)  3H2 (g) + CO (g)
2. For the following unbalanced reaction, there are 5.00 g. of each reactant.
Determine which reactant is limiting and determine the mass of each of the
products:
H2S (g) + O2 (g)  SO2 (g) + H2O (l)