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### Limiting Reactant

Posted: Fri Oct 05, 2018 12:56 am
When doing limiting reactant problems I have been told to convert the given samples of reactants (in grams ) to moles separately and calculate two separate theoretical yields using stoichiometry and see which one produces less product in order to determine the limiting reactant. Is there a more efficient way to approach these problems where you do not have to carry out two separate full yield calculations or is this the only way to do it?

### Re: Limiting Reactant

Posted: Fri Oct 05, 2018 12:57 am
I'm pretty sure you need to do both calculations.

### Re: Limiting Reactant

Posted: Fri Oct 05, 2018 1:03 am
I don't believe you need to find theoretical yield for limiting reactant problems. To find the limiting reactant, just take the information you're given and convert to moles (however needed). Once you have all the reactants in moles, compare to see which uses more moles quickly based on the given amounts of reactants you have.

### Re: Limiting Reactant

Posted: Fri Oct 05, 2018 1:14 am
In regards to a limiting reactant problem, you won't always be asked to calculate percent yield, as it may just be asking you to determine whether a limiting reactant is present. However, if asked to find percent yield, the most efficient way in my opinion is as follows:

1. Balance equation
2. Calculate molar mass of reactants and products
3. Convert given masses to moles
4. Determine limiting reactant by comparing calculated moles to required moles (check balanced equation)
5. Calculate moles of product that can form using limiting reagent
6. Convert the moles of the products to grams (using molar mass) to calculate theoretical yield

percent yield= actual yield/theoretical yield x 100

### Re: Limiting Reactant

Posted: Fri Oct 05, 2018 1:25 am
Instead of calculating each possible theoretical yield, you can also just calculate the moles of each given substance and compare it to the number of moles in the balanced equation. The substance with the smallest ratio of [moles you have/coefficient in the balanced chemical equation] is the limiting reactant.

So for example:
CaC2 + 2H2O —> Ca(OH)2 + C2H2
What is the amount of C2H2 produced given 100g of CaC2 and 100g of H2O?

***convert the grams of each substance to moles

100 g CaC2 (1 mole CaC2/64.1 g CaC2) = 1.56 mole CaC2
100g H2O (1 mole H20/18 g H20) = 5.55 mole H2O

***now compare the moles you have to the moles "required" by the balanced equation. The substance that has a smaller ratio of moles you have to moles required is the limiting reactant. In other words, divide the moles you have by the coefficient in front of that substance in the balanced equation.

1.56 mole CaC2/1 = 1.56
5.55 mole H2O/2 = 2.78

***since 1.56 is less than 2.78, the ratio of "moles you have" vs "moles required" is smaller for CaC2. So CaC2 is the limiting reactant

From here you can convert from moles of one substance to moles of the product and finally to grams of the product.
1.56 moles CaC2 (1 mol C2H2/1 mol CaC2) (26 g C2H2/1 mole C2H2) = 40.6 g C2H2

### Re: Limiting Reactant

Posted: Fri Oct 05, 2018 8:22 pm
Hii! For me, I calculate the reactant that is the limiting reactant first, then I use the number of moles of the limiting reactant to calculate that of the products. I think the reason for this is because the products produced cant be "greater"than that of the limiting reactants. (But correct me if Im wrong). Hope that helps!

### Re: Limiting Reactant

Posted: Sat Oct 06, 2018 6:21 pm
Instead of calculating both theoretical yields, one can use the mole ratios in the balanced chemical equation to determine which substance is the limiting reactant. Of course this may only be completed after the amount in moles for the substances are converted from grams or otherwise determined.

### Re: Limiting Reactant

Posted: Sun Oct 07, 2018 7:41 pm
I think finding the theoretical yield for both reactants that are considered is much easier to get a grasp for what is the limiting reactant but I was curious if there was a faster way that always works.

### Re: Limiting Reactant

Posted: Sun Oct 07, 2018 9:51 pm
I believe that you can instead just convert the grams to moles of each reactant/product. Then, you can use the mole ratios of the reactants and products to determine which is the limiting reactant. For example, for the equation CaC2 + 2H2O = Ca(OH)2 + C2H2, the reactants have a mole ratio of 1:2, so you can use this to determine whether the amount of moles for which reactant is in excess, and thus the other reactant is the limiting reactant.