Need help preparing for the General Chemistry section of the MCAT? MedSchoolCoach expert, Ken Tao, will teach everything you need to know about Balancing Equations, Limiting Reagent, and Theoretical Yield for Stoichiometry. Watch this video to get all the MCAT study tips you need to do well on this section of the exam!
Let's discuss how to balance chemical equations, how to determine the limiting reagent of a reaction, and how to calculate the theoretical yield and actual yield of a reaction. Let’s begin with balancing an equation. The goal of balancing equations is to ensure that you have the same number of atoms for each element on both sides of the reaction. This ensures that we have written out the correct reaction, and that our reaction formula will accurately predict the stoichiometric ratios of reactants and products. Remember that mass, and charge, are conserved, so however many atoms of a particular element we begin with, we must also end with.
Balancing Equations
As an example, let’s consider the following reaction, where we have propane reacting with oxygen to form carbon dioxide and water.
C3H8 + O2 → CO2 +H2O
Let’s begin by tallying up the number of atoms of each element we have in our reactants, in this case three carbon atoms, eight hydrogen atoms, and two oxygen atoms. Next, repeat the same process for the product compounds, on the right side of the equation. The products contain one carbon atom, two hydrogen atoms, and three oxygen atoms, an unbalanced equation. To balance the equation, we will first pick one atom to balance. The easiest atom to balance will typically be an atom that is present in its elemental state (in this case, oxygen) so we will save that as the last atom to balance. We could start with either carbon or hydrogen. In this case, let’s begin with carbon.
The imbalance of carbon is weighted to the left side of the reaction. Therefore, to balance it, we must add a coefficient to the product containing carbon, CO2. If we had a ‘3’ coefficient before carbon dioxide, carbon will now be balanced, giving us this intermediate equation:
C3H8 + O2 → 3CO2 +H2O
Remembering that we’ll save oxygen for last, let’s now balance hydrogen. Hydrogen is also imbalanced to the left side of the equation, meaning a coefficient will need to be added to the products. There are two hydrogen atoms in water and eight in propane, so balancing hydrogen will require adding a ‘4’ coefficient before water. Our not-yet balanced equation will now look like this:
C3H8 + O2 → 3CO2 +4H2O
Our final step will be to balance oxygen. There are currently two oxygens on the left side of the equation and, with our new balancing coefficients, ten oxygens on the right side of the equation. Therefore, adding a ‘5’ coefficient before elemental oxygen will balance the oxygens across the reaction. This will give us a final balanced equation of:
C3H8 + 5O2 → 3CO2 +4H2O
We can see that there are now three carbons, eight hydrogens, and ten oxygens on the reactant side, as well as three carbons, eight hydrogens, and ten oxygens on the product side. Reflecting back on our discussion of stoichiometric ratios, this means that the combustion of one mole of propane will require five moles of oxygen, and will produce as products three moles of carbon dioxide and four moles of water (and quite a bit of heat, not shown in the reaction).
Limiting Reagent
However, this reaction formula assumes that we have exactly a five to one ratio between the amount of available oxygen and propane. In real-world chemical reactions, the relative ratios of reactants do not typically exactly match their stoichiometric ratios. In these instances, one reactant will be entirely consumed when some amount of other reactants are still remaining, halting the reaction. The reactant that will be the first to be consumed completely is called the limiting reagent. For example, in the above reaction, if only three moles of oxygen were provided to one mole of propane, the reaction would halt after the three moles of oxygen were consumed, with some amount of propane (specifically, 2/5th of a mole of propane) left over. Additionally, the amount of carbon dioxide and water produced would be less than three and four moles, respectively.
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