How to find the limiting reactant easily and quickly

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Jessica Dharmawan 1G
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Joined: Fri Sep 28, 2018 12:15 am

How to find the limiting reactant easily and quickly

Postby Jessica Dharmawan 1G » Mon Oct 08, 2018 10:45 pm

I'm having trouble finding easy ways to determine the limiting reactant. Any advice???

There is also a problem in one of the post-assessment modules that asks questions about the chemical equation 2A + 1B --> 3C, such as what reactant is in excess and what is the maximum amount of product that can be produced, and I'm having a hard time understanding how to solve these problems.

katherinemurk 2B
Posts: 68
Joined: Wed Nov 15, 2017 3:02 am

Re: How to find the limiting reactant easily and quickly

Postby katherinemurk 2B » Mon Oct 08, 2018 11:02 pm

Okay first balance the equation given if it is not already. Then if given the mass of the two reactants convert both to moles. Then divide each of those amounts by the number of moles in the equation.
For example
2Al+6HCI --> 2AlCI3 + 3H2
If given the grams of Al and HCI you convert to moles. In this case there are 5.43 g of Al and 7.80 g of HCI. When you convert to moles you get .201 mol Al and .214 mol HCI. Then divide each number by the number of moles in the equation. So .201 mol Al/2 and .214 mol HCI/6. The smallest amount is the limiting reactant. If you divide, you would see the limiting reactant is HCI. Hope I didnt confuse you

juliasloan_4g
Posts: 67
Joined: Fri Sep 28, 2018 12:28 am

Re: How to find the limiting reactant easily and quickly

Postby juliasloan_4g » Mon Oct 08, 2018 11:33 pm

I've always found that the easiest way to find the LR(but maybe not the fastest) is converting both reactants into one product. You do this by taking the mass given to you of both products and using molar mass and molar ratios to convert into product. You can convert to either moles of grams, both work. Whichever reactant produced a lesser amount of the product is the limiting reactant. I prefer this method because it often leads you to later problems as well such as how much of a product is actually produced(you know because its whichever result is from the LR).

Saman Andalib 1H
Posts: 73
Joined: Fri Sep 28, 2018 12:16 am

Re: How to find the limiting reactant easily and quickly

Postby Saman Andalib 1H » Tue Oct 09, 2018 1:55 am

Think about simple ratios when dealing with limiting reactant questions. The easiest way is to convert both reactant quantities into moles of the particular product that we are solving for. This will allow you to easily observe which one of the reactants produces the least amount of product and is, therefore, the limiting reactant.

Gary Qiao 1D
Posts: 66
Joined: Fri Sep 28, 2018 12:26 am

Re: How to find the limiting reactant easily and quickly

Postby Gary Qiao 1D » Tue Oct 09, 2018 2:24 am

What I usually do is first make sure that the chemical equation is balanced, and then I try to convert the grams of the reactants to moles (unless the problem already gives the number of moles of each reactant). Then, I would take these mole values and divide them by the stoichiometric coefficients in the chemical equation. From there, whichever reactant has the least value would have to be the limiting reactant. You would then just use the limiting reactant amount in order to find the maximum product (so pretty much disregard the excess reactants when finding the maximum amount of product that can be produced).

Courtney McHargue 1I
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Joined: Fri Sep 28, 2018 12:17 am

Re: How to find the limiting reactant easily and quickly

Postby Courtney McHargue 1I » Wed Oct 10, 2018 11:49 am

When I find the limiting reactant I always use the molar masses of the reactants to find how many moles of each the reaction has. After I do that I look at the molar ratios between the reactants to see how much of each is needed to react with the other. In the example 2A + 1B --> 3C, for every 2 moles of A, you only need 1 mole of B. So, if you find that you have 1.00 mol of both A and B you know that A is the limiting reactant, as for one mole of B to react completely you would need 2 moles of A, and since the ratio is 2:1, with the 1 mole of A that you have you only need 0.5 mol of B, showing that A is the limiting reactant and that B will be in excess at the end of the reaction.


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