Limiting Reactants With Balanced Equations

Limiting reactant problems that are solved using balanced equations are very much the same as what you were working with in the previous section. It's just that you now use the balanced equation to give you the relationship between the chemicals.

There is a sequence of steps that should be taken to solve these types of problems. First, write an equation for the reaction and balance it. Second, determine the mole and weight relationships among the chemicals in the reaction. Third, determine the limiting reactant in the same manner you did in the previous section. Fourth, carry out the necessary calculations using the mole and weight relationships determined from the balanced equation.

Example 1: Masses From Balanced Equation

Methane (CH₄) reacts with oxygen (O₂) to form carbon dioxide (CO₂) and also water vapor (H₂O).

If 20.0 g of CH₄ was mixed with 30.0 g of O₂ and ignited, what would be the limiting reactant? How much excess reactant would be left over? How much CO₂ and H₂O would be produced?

We will need to go through a sequence of steps to solve this problem. So, the first thing we need is to write down the equation for the reaction and balance it (below). We will also need to determine the weight relationships. Everything in the question is in grams, so we don't really need to write down the mole relationship, but it helps in the next step.

We need the weight relationships, which are shown. A very useful thing at this point, just to double check your math is to realize that the conservation of mass applies. So, the total weight of reactants should be equal to the total weight of the products. Checking that will serve as a double check on the math that you've done so far. Sixteen plus 64 is 80 and 44 plus 36 is also 80, so that checks out.

CH₄ + 2 O₂ → CO₂ + 2 H₂O

1 mol 2 mol 1 mol 2 mol

16.0 g 64.0 g 44.0 g 36.0 g

Now, we need to determine the limiting reactant in the same way that you did earlier in this lesson. In this case that can be done using a rough approximation.

We have 20.0 g of methane mixed with 30.0 g of oxygen. What we have available is half-again as much oxygen as we have methane. However, if you look at the weight relationship, you can see that it takes about four times as much oxygen. So we're going to run out of oxygen. We just don't have enough oxygen to react with all of the methane that's available. The limiting reactant is 30.0 g of oxygen.

Using the weight relationship of CH₄ to O₂, we can see that with the amount of O₂ we are given, 7.5 g of CH₄ would be required to react fully with it. Since we have 20.0 g of CH₄,there is more than enough, so CH₄ is the excess reactant and oxygen is limiting. This verifies our approximation from before, and also allows us to answer the second part of the question: how much excess reactant would be left over. Starting with 20.0 g CH₄ and using 7.5, we would be left with 12.5 g of excess reactant.

As shown above, this reaction would make 20.6 g of CO₂. You should set up the corresponding calculation for H₂O and verify that there would be 16.9 g of water produced.

Example 2: Moles From Balanced Equation

P₄ and O₂ react to make P₂O₅. Balance the equation for this reaction, and determine the limiting reactant if 4.5 moles of P₄ are mixed with 16.7 moles of O₂. Find the mass of product made.

Once again, we will need a balanced equation and weight/mole relationships to solve the problem. The equation is fairly straightforward to balance, and the relationships are shown below.

P₄ + 5 O₂ → 2 P₂O₅

1 mol 5 mol 2 mol

123.9 g 160.0 g 283.9 g

As before, we can start things off with a rough estimate. Looking at the ratios from our table, you can see that we need five moles of oxygen for every mole of phosphorus. Do we have that in this case? It may or may not be clear, depending on how good you are at multiplying decimals in your head. Either way, let's do the actual math to see whether we have enough oxygen to use up all our phosphorus.

Our calculations reveal that using up all 4.5 moles of phosphorus would require 22.5 moles of oxygen. Since we only have 16.7 moles of oxygen, we do not have enough, making oxygen limiting and phosphorus the excess reactant.

We are also asked to find the mass of product formed in this reaction. Since we have determined this to be oxygen, we will use our 16.7 moles of oxygen as the starting point, and a mixed grams-moles stoichiometric ratio as a conversion factor. The calculation reveals that 948 g of product would be made in this situation.

Your lab workbook contains a number of practice problems that will help you get comfortable with limiting reactant calculations. It is strongly recommended you work on these, as these problems come in many different flavors. Ask your instructor if you are running into trouble.