## Determining Limiting Reactant

LMendoza 2I
Posts: 41
Joined: Sat Jul 22, 2017 3:00 am

### Determining Limiting Reactant

Does anyone know if the limiting reactant will always be the product with the least amount of moles but when converted into grams, will the LR be the product with the greatest amount of theoretical grams?

Nora 1F
Posts: 46
Joined: Fri Sep 29, 2017 7:04 am

### Re: Determining Limiting Reactant

The limiting reactant will always be the reactant that runs out first, thereby making it so that the reaction cannot take place any longer. Sometimes it will be the reactant with the least amount of moles, but many times it will not be. In order to determine which reactant is the limiting reactant, you first have to find the amount of moles of each reactant. Then you use the balanced chemical equation of the reaction to find the molar ratio between the reactants. Using the coefficients, you should be able to compare the amount of moles of the reactants you HAVE with the amount that you NEED according to the equation.
For example, if your equation is 1A + 2B ---> 1C and you have 0.500 moles of reactant A and 0.750 moles of reactant B, you will use the ratio of 1:2 to determine which reactant is in excess. Using the ratio, we know that if we have 0.500 moles of reactant A, we need twice as much of reactant B, or we need 1.0 mole of reactant B. Because we only have 0.750 moles of reactant B (and even though this is a greater amount than 0.500), we can say reactant B is the limiting reactant. Reactant B will determine how much product is produced because once those 0.750 moles are used up, the reaction won't be able to continue even if there is still some of reactant A left.
In order to find out how many moles of product C would theoretically be formed, you would use the molar ratio between the limiting reactant and product C, 2:1, and the amount of moles of reactant B. This would mean that there should be half as many moles of C as there are of B. Converting this number (0.375 mol C) to grams will give you the theoretical yield of C.

I did not completely understand the wording of your question, but I hope this helps!

Alexa_Henrie_1I
Posts: 61
Joined: Fri Sep 29, 2017 7:03 am

### Re: Determining Limiting Reactant

Just to clarify, when determining molar mass of a reactant you don't include the coefficient that was found when balancing the equation right?

Kathleen Vidanes 1E
Posts: 62
Joined: Fri Sep 29, 2017 7:07 am

### Re: Determining Limiting Reactant

If given the grams of only one reactant within a question, should we automatically assume that that particular reactant is limiting?

Sabah Islam 1G
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

### Re: Determining Limiting Reactant

When determining the molar mass of a reactant, you don't have to include the coefficient that was found when balancing the equation. You use the stoichiometric coefficient when using dimensional analysis to figure out how many moles of a certain reactant is equivalent to the moles of another reactant or product. If given the grams of only one reactant in a question, you wouldn't necessarily assume that it is the limiting reactant unless the question states that another reactant is in excess, but there is also a possibility of that reactant being limiting. Not sure if I answered your question, but I hope this helps!

soniatripathy
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: Determining Limiting Reactant

I find that it is easier to calculate the limiting reactant by dividing the moles of each reactant by its stoichiometric coefficient. This makes is easier to see that the smaller quotient is the limiting reactant.

Gurkriti Ahluwalia 1K
Posts: 37
Joined: Fri Sep 29, 2017 7:07 am

### Re: Determining Limiting Reactant

sometimes, the limiting reactant is not always the one with the least value given. You should always use the ratios present in the chemical equation because although there may be less of one reactant you may need more than what's provided of the other. for example, if a ratio is 1:5 and you are given 15 moles of the first reactant and 25 moles of the second, it is actually the SECOND reactant that is limiting, because if the ratio is 1 to 5, then you will need 75 moles of the second reactant.