General Chemistry Limiting Reagent

Question

The combustion of propane C₃H₈ is shown below

C₃H₈(g)  +  5O₂(g) →  3CO₂(g)  +  4H₂O(g) 

If this reaction was performed with 3 moles of propane and 10 moles of oxygen, Identify the limiting reagent

Answer 1

Hi, Joselyne,

The "secret" to solving this problem is recognizing that the ratios provided in the equation are CONSTANT (meaning that for this particular reaction, those ratios will always be as shown here):

C₃H₈(g)  +  5O₂(g) →  3CO₂(g)  +  4H₂O(g) 

So, the ratio of propane to oxygen is (and will be!) 1:5. Therefore, if you are performing a reaction with 3 moles of propane and 10 moles of oxygen, you can solve for the limiting reagent as follows:

With 3 moles of propane, the overall molar quantities used/produced would be:

3×1 C₃H₈(g)  +  3×5 O₂(g) →  3×3 CO₂(g)  +  3×4 H₂O(g) ,

or 3 C₃H₈(g) +  15 O₂(g)  →   9 CO₂(g)   +   12 H₂O(g).

As you can see, to fully consume 3 moles of propane would require 15 moles of oxygen to satisfy the combustion equation ... but you don't have 15 moles of O₂(g), you only have 10! Therefore, oxygen is the limiting reagent in this reaction.

If you want to check your results, we can solve the same equation starting with the 10 moles of oxygen. In this case:

2×1 C₃H₈(g)  +  2×5 O₂(g) →  2×3 CO₂(g)  +  2×4 H₂O(g) 

2 C₃H₈(g)  +   10 O₂(g)  →  6 CO₂(g)  +  8 H₂O(g) 

Here, you see that 10 moles of oxygen would consume 2 moles of propane ... and there would still be one mole of propane remaining (in excess). The limiting reagent - the one to be exhausted first - would be oxygen.

Thanks,

Chris

Answer 2

For a limiting reagent question, you essentially want to figure out which reactant will run out first. This should be the same for both products, but it never hurts to double check. Let's start with the amount of CO₂ that can be created from each reactant. Because they are given to you in moles, you don't have to calculate the molecular weights, you are just using the mole ratios. For example, you will create 3 moles of CO₂ from 5 moles of O₂.

3 mol propane (C₃H₈) x (3 mol CO₂ / 1 mol propane) = 9 mol CO₂

10 mol O₂ x (3 mol CO₂ / 5 mol O₂) = 6 mol CO₂

The limiting reagent is O₂ because you will create less CO₂. Therefore, the O₂ would run out first and you'd only have created 6 mol CO₂, with some propane left over.

Let's check our answer by determining the limiting reagent for water creation.

3 mol propane x (4 mol H₂O / 1 mol propane) = 12 mol H₂O

10 mol O₂ x (4 mol H₂O / 5 mol O₂) = 8 mol H₂O

As you can see, the limiting reagent is still O₂. You can follow this same format for any problem, as long as you use the mole ratio of products to reactants.