Finding Excess Amount of Reactant  [ENDORSED]

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Melissa_Aguirre1J
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Joined: Tue Nov 14, 2017 3:02 am

Finding Excess Amount of Reactant

Postby Melissa_Aguirre1J » Sun Apr 15, 2018 10:11 am

How do you find the excess amount of the reactant that is not the limiting reactant in a limiting reactant problem?

Alejandro Salazar 1D
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

Re: Finding Excess Amount of Reactant

Postby Alejandro Salazar 1D » Sun Apr 15, 2018 10:52 am

When you convert g× mol/g ×mol/mol twice to determine the limiting reactant, you choose the reactant in moles with the lowest value since that's the one that will "run out" first

For Excess reactant it is similar, but instead choose the higher value that is in moles. Depending on the question, you will probably need to convert that value in moles into grams and subtract the grams from the original value of grams given in the problem.

Anna De Schutter - 1A
Posts: 66
Joined: Wed Feb 21, 2018 3:01 am

Re: Finding Excess Amount of Reactant  [ENDORSED]

Postby Anna De Schutter - 1A » Sun Apr 15, 2018 4:23 pm

When you are trying to determine the limiting reactant of the reaction, you find the initial amount of moles for each reactant (either given to you or you convert masses to moles using the molar mass) and then you choose one reactant (let's call it reactant 1) and depending on the molar ratios (given by the stoichiometric coefficients) you determine the amount (in moles) needed of the other reactant (let's call it reactant 2) for this reaction to go to completion.

If you have more of what you need of reactant 2 then the limiting reactant is reactant 1. The reactant in excess is then reactant 2 and the amount of moles used up in the reaction is the one calculated thanks to the molar ratios. So amount of reactant 2 left = total amount of reactant 2 - amount of reactant 2 needed for this reaction to go to completion. All of this is in moles, you can then convert it to grams using the molar mass if need be.

If you have less of what you need of reactant 2 then the limiting reactant is reactant 2. The reactant in excess is then reactant 1 and the amount of moles used up in the reaction can be calculated using the molar ratios and the initial amount in moles of reactant 2 we have. So amount of reactant 1 left = total amount of reactant 1 - amount of reactant 1 needed for this reaction to go to completion. All of this is also in moles, you can then convert it to grams using the molar mass.

I hope this helps! :)
Anna De Schutter - section 1A


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