# Limiting Reactants Ok, I know Stoichiometry can be tricky enough without adding an extra twist to it, but lets talk about limiting reactants. I know I know, you probably feel like our buddy Brick does but just here me out. Limiting reactants are elements or molecules that limit the amount of product that can be produced in a reaction. If that doesn’t quite make sense, let me explain it like this…

Lets say you are a luxury car manufacturer. Like, big time luxury. Like, handmade cars… leather, wood, gold trim, the works. And for the sake of argument, and this example, that at the end of production the last thing your company does is put the tires on the finished car bodies. Now the end of production would follow this reaction:

# → So if I had 16 tires and 4 car bodies I could make…    ta-da, 4 cars.

BUT! What if my tire supplies were running low and I only had 8 tires and 4 car bodies. How many cars could I make then?

# I could only make 2 cars!

The amount of tires I had, limits the amount of cars I can produce (important word). Therefore, my tires are my limiting reactant! Because they limit the amount I can produce.

Now, lets apply that same concept to a chemical reaction, using what we have learned about Stoichiometry and it should be no sweat! First, we need a balanced chemical reaction. Then we need a given. Lets do the first example from our Limiting reactants packet as our reaction.

 Example: 3 O2 + CS2 → CO2 + 2 SO4 Given: 5.55 moles 2.45 moles

So, if you notice, we have 2 givens. These would be the amounts we put together in a reaction… BUT! we have to figure out which one is the limiting reactant! So, first lets assume O2 is the LR and see how much product (CO2) we can produce…

 5.55 mole O2 1 mole CO2 3 mole O2 = 1.85 mole CO2

So we’ll put our new information into the chart to keep track of our data. It would look something like this…

 Example: 3 O2 + CS2 → CO2 + 2 SO4 Given: 5.55 moles 2.45 moles Assume O2is LR 5.55 moles 1.85 moles

Now lets assume CS2 is the LR and see how much product (CO2) we can produce…

 2.45 mole CS2 1 mole CO2 1 mole O2 = 2.45 mole CO2

Now our chart will look something like this…

 Example: 3 O2 + CS2 → CO2 + 2 SO4 Given: 5.55 moles 2.45 moles Assume O2is LR 5.55 moles 1.85 moles Assume CS2is LR 2.45 moles 2.45 moles

In conclusion the reactant that produces the least amount of product is the 5.55 moles of O2! It only produced 1.85 moles of CO2, while 2.45 moles of CS2 produces 2.45 moles!
So O2 is our Limiting reactant and CS2 is our Excess Reactant!

Now, there are some further steps you’ll have to do in the packet (how much Excess Reactant was used, what amount is leftover, what are the moles of the other product) and you may have a given in grams that you’ll have to convert… but all these are things you have done before. When in doubt, read the question carefully. There are key words that will help you determine your given and what you are solving for! Good Luck!