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Let’s say that we have this equation, which is the combustion of methane. Then I do the same on the right side: one carbon, two hydrogens, and three oxygens. The only way to get that four hydrogens is to double the amount of …and this equation is now balanced.You’ve probably learned much of this even in a non-AP chemistry class, so bear with me; it’ll get more advanced.
However, if this seems obvious to you, keep it in mind; it may come in handy at some point.
Titration may seem like an intimidating topic, but while it is a little involved, it simply builds on the concepts we already know from expected values and limiting reactants. A little explanation: titration is basically just adding one compound to another in solution (in this case, adding a base to an acid) and seeing how the p H changes. Those are the five main applications of stoichiometry on the AP Chemistry exam.
First, let’s talk about the basic concepts of stoichiometry.
First of all, we have to start with a balanced reaction. I add up the number of atoms of each element on the left side: one carbon, four hydrogens, and two oxygens. First, we look at the left side and see that we need four hydrogens.
You see, the conversions from liters to moles and back to liters are the same for both gases, so the only number that really mattered, in this case, was the mole ratio between the reactant and the product (in other words, the fact that it took two . Well, first of all, it’s good practice to do it the long way, and it helps to understand the fundamentals.
And on the AP Chemistry exam, it’s often worth the extra two minutes to write out your entire thought process on a stoichiometry problem, especially if it’s a long answer.
The AACT high school classroom resource library has everything you need to put together a unit plan for your classroom: lessons, activities, labs, projects, videos, simulations, and animations.
We constructed a unit plan using AACT resources that is designed to teach the concepts of stoichiometry and limiting reactants to your students.
Sadly, for this example, we have to move on from our beloved methane combustion equation ( Pretty intimidating if you don’t know what to do, right? In this case, we can see that the p H passes through the equivalence point (where the acid and the base cancel each other out and the p H is of oxalic acid. You’ll notice that the common thread running through all of them is the expected value.
Essentially, if you understand expected value really well, you should be able to figure out most of the rest.