How To Solve Mass To Mass Stoichiometry Problems

This concept is very important to the applications of stoichiometry. For the sake of simplicity, we’re going to continue to use the combustion of methane as our reaction in question. We have more moles of oxygen, so methane is the limiting reactant, right?No, because in the reaction, we can see that we need twice as many moles of oxygen as of methane.However, if this seems obvious to you, keep it in mind; it may come in handy at some point.

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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.

This balanced equation tells us a lot about how this reaction works. This one is another very important application of stoichiometry for the AP Chemistry exam.

The core concept we can take away from it is that one molecule of methane combines with two molecules of diatomic oxygen to form one molecule of carbon dioxide and two molecules of water, or correspondingly, that one mole of methane combines with two moles of diatomic oxygen to form one mole of carbon dioxide and two moles of water. A note about nomenclature: saying reagent makes me feel like a mysterious alchemist, but I’m going to favor reactant because I think it’s a little bit clearer.

That being said, if you’re pressed for time, you can try it.

Recall this question: , meaning that we need twice as many moles of oxygen as we do methane.

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.

The standard practice to eliminate this problem is to simply carry out both calculations as expected value calculations.

So, let’s pretend that we’re calculating the expected value of , so oxygen is the limiting reactant here because its expected value in the reaction was less.


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