Different Way to Find the Limiting Reactant

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Ariel Fern 2B
Posts: 105
Joined: Fri Aug 30, 2019 12:17 am

Different Way to Find the Limiting Reactant

Postby Ariel Fern 2B » Sun Oct 06, 2019 10:15 pm

Hi everyone! I just wanted to share a different way that my high school chem teacher taught me how to discover which reactant would be limiting. Let me know if there are any pros or cons to this method. I'll use M5 as an example.

The reaction 6 ClO2(g) + 2 BrF3(l) ---> 6ClO2F(s) + Br2(l) is carried out with 12 mol ClO2 and 5 mol BrF3.

We were taught to divide the number of moles available by the stoichiometric coefficient of the respective reactant:
12 mol ClO2 / 6 = 2
5 mol BrF3 / 2 = 2.5

Since the ratio for ClO2 is smaller, that means that ClO2 is the limiting reactant and BrF3 is the excess reactant. Whichever is the least is the limiting reactant.

Hope this helps! It has helped me to double check to make sure I am on the right track!

Petrina Kan 2I
Posts: 102
Joined: Fri Aug 30, 2019 12:17 am

Re: Different Way to Find the Limiting Reactant

Postby Petrina Kan 2I » Sun Oct 06, 2019 11:27 pm

Thank you this is actually very helpful! Just curious, but I don't believe this method works if the problem just gives you the number of grams of each reactant? (Or it just wouldn't be as simple and straightforward)

ishaa Diwakar 4E
Posts: 56
Joined: Wed Sep 11, 2019 12:17 am

Re: Different Way to Find the Limiting Reactant

Postby ishaa Diwakar 4E » Sun Oct 06, 2019 11:31 pm

I actually learned this method in high school too! It is actually similar to one of the methods Lavelle taught, where he told us to solve for the amount of products each molar amount of the reaction would form. This method basically just removes the second step of solving for the products.

Ariel Fern 2B
Posts: 105
Joined: Fri Aug 30, 2019 12:17 am

Re: Different Way to Find the Limiting Reactant

Postby Ariel Fern 2B » Mon Oct 07, 2019 5:01 pm

Hi Petrina and Ishaa!

I'm glad to hear that this has helped and is another viable option for you! If the problem gives the quantities in grams, then I convert grams into moles by dividing by molar mass, which then allows me to use the method! Hope that makes sense!


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