Limiting Reactant

Husnia Safi - 1K
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Limiting Reactant

Hi can someone please explain what ratios have to do with finding the limiting reactant. Thanks!

Jessica Schirmer 1J
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Re: Limiting Reactant

Once you calculate the moles of each reactant you have, given that the problem gave you two masses to begin with, you compare the ratios of those moles to the molar ratio provided by a balanced chemical equation. (The molar ratio is given by the stoichiometric coefficients).

For example, if the ratio you calculated for reactants X and Y was 1.68 : 2.4, but the balanced chemical equation has a ratio of 1 : 3, you therefore know that the limiting reactant was element Y because 2.4 is not three times as much as 1.68.

Chem_Mod
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Re: Limiting Reactant

You have to use the ratios of the reactants from the balanced chemical equation to find which reactant is the limiting reactant. Essentially you can think of a balanced chemical equation like a cooking recipe. If you need 2 cups of water and 1 cup of sugar to make something and you only have 2 cups of each, then water is your limiting reactant, and you will have an excess of 1 cup of sugar, because you only need 1 cup of sugar for every 2 cups of water. This concept is analogous in finding limiting reactants in chemistry. If 2A + 3B --> A2B3, then you need 3B for every 2A (or 2A for every 3B), so you determine which reactant will limit how much A2B3 you will produce. As a reminder, note that the coefficients in balanced chemical equations are in moles, so make sure you convert information you are given to moles if they are not in moles (i.e., grams).

Minie 1G
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Re: Limiting Reactant

When calculating for the limiting reactant, you need to not only find how many moles of each (cause usually you're given the reactants in grams of mass) you have, but how many moles of one reactant is needed in relation to the other reactants.
For example, if your equation is X + 3Y -> 2Z then every 1 mole of X would need to react with 3 moles of Y to produce 2 moles of Z. The molar ratio of X to Y, is 1:3. You need 3 times as many moles of Y as you do of X.

Raquel Floyd 1K
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Re: Limiting Reactant

Could someone explain in more detail the "comparing" part of these steps? I do not understand what you do after you have the two different ratios, how do you compare them to know which is your limiting reagent?

Allen Chen 1J
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Re: Limiting Reactant

Raquel Floyd 1K wrote:Could someone explain in more detail the "comparing" part of these steps? I do not understand what you do after you have the two different ratios, how do you compare them to know which is your limiting reagent?

Here is an example:
2Na+Cl2 = 2NaCl.
If 8 g of Na reacts with 8 g of Cl2:
you get .349 mol of Na and .114 mol of Cl.
the ratio of these is .349/.114.
However, the ration of Na to Cl2 in the equation above is 2/1.
Since .349/.114 is greater than 2/1, Na is in excess and therefore Cl2 is the limiting reactant.

Isabel Jabara 1C
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Joined: Fri Apr 06, 2018 11:03 am

Re: Limiting Reactant

Raquel Floyd 1K wrote:Could someone explain in more detail the "comparing" part of these steps? I do not understand what you do after you have the two different ratios, how do you compare them to know which is your limiting reagent?

Using Allen's example, if you have .349 mol of Na, and the ratio of Na to Cl2 is 2:1, then .349 mol of Na requires .349x2 mol of Cl2, or .968 mol Cl2. You only have .114 mol of Cl2 but you need .968 mol, so Na is excess and Cl2 is the limiting reactant.

Paywand Baghal
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Joined: Fri Apr 06, 2018 11:01 am

Re: Limiting Reactant

Jessica Schirmer 1J wrote:Once you calculate the moles of each reactant you have, given that the problem gave you two masses to begin with, you compare the ratios of those moles to the molar ratio provided by a balanced chemical equation. (The molar ratio is given by the stoichiometric coefficients).

For example, if the ratio you calculated for reactants X and Y was 1.68 : 2.4, but the balanced chemical equation has a ratio of 1 : 3, you therefore know that the limiting reactant was element Y because 2.4 is not three times as much as 1.68.

Would you have to always base the limiting reactant on the first part of the ratio. So for this question, do you base the 2.4 on 1.68 meaning Y is the limiting reactant. I dont know if my question makes sense, or because 1.68 is not 1/3 of 2.4, then X is the limiting reactant?

Isabel Jabara 1C
Posts: 34
Joined: Fri Apr 06, 2018 11:03 am

Re: Limiting Reactant

Paywand Baghal wrote:
Jessica Schirmer 1J wrote:Once you calculate the moles of each reactant you have, given that the problem gave you two masses to begin with, you compare the ratios of those moles to the molar ratio provided by a balanced chemical equation. (The molar ratio is given by the stoichiometric coefficients).

For example, if the ratio you calculated for reactants X and Y was 1.68 : 2.4, but the balanced chemical equation has a ratio of 1 : 3, you therefore know that the limiting reactant was element Y because 2.4 is not three times as much as 1.68.

Would you have to always base the limiting reactant on the first part of the ratio. So for this question, do you base the 2.4 on 1.68 meaning Y is the limiting reactant. I dont know if my question makes sense, or because 1.68 is not 1/3 of 2.4, then X is the limiting reactant?

You could do it either way. If you wanted to base it on the second number, you would calculate the ratio of 1X:3Y with the 3Y being 2.4. Therefore, the first number in the ratio should be a third of 2.4, which is 0.8 (so your ratio is 1:3=0.8:2.4). Since you are given 1.68 of X and you need 0.8 of X to have 2.4 Y, then you have excess X so Y is still the limiting reactant.

004985802
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Re: Limiting Reactant

using molar mass you can find the ratio of moles you are able to use (max) of a limiting reactant in a chemical equation to the amount in the equation, and apply that ratio to other chemicals/compounds to see how much of their total molar mass will be consumed in the reaction (and use their molar masses to calculate the amount in grams used)