Steps to calculate limiting reagant
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 Posts: 58
 Joined: Wed Sep 18, 2019 12:22 am
Steps to calculate limiting reagant
When solving for the limiting reactant, what step follows directly after calculating moles of each molecule? Also can you explain the next few steps and why they matter?
Re: Steps to calculate limiting reagant
(1) Find the moles of each reactant. Converting the mass to moles will allow easier comparison between different reactants.
(2) Use molar ratios in order to calculate the required moles needed for the reaction to occur.
(3) Compare the required moles to the available moles (from step 1) to determine the limiting reactant. The limiting reactant will be the reactant in which the moles in Step 2 is higher than the moles in Step 1. Determining the limiting reactant will inherently help determine how much product can be made in the reaction given the amount of reactants available.
(2) Use molar ratios in order to calculate the required moles needed for the reaction to occur.
(3) Compare the required moles to the available moles (from step 1) to determine the limiting reactant. The limiting reactant will be the reactant in which the moles in Step 2 is higher than the moles in Step 1. Determining the limiting reactant will inherently help determine how much product can be made in the reaction given the amount of reactants available.

 Posts: 101
 Joined: Wed Sep 18, 2019 12:20 am
Re: Steps to calculate limiting reagant
After you calculate the moles of each molecule involved in the reaction, you must compare the ratio between quantities of the different reactants to the ratio of their coefficients in the balanced equation for the chemical reaction. For instance, in the chemical equation 2NA+2H20>2NaOH+H2, there is a 1:1 ratio between sodium and water. Therefore, theoretically, if there is an equal number of moles of each reactant, neither will be the limiting reactant since every atom of sodium will have a molecule of water to react with; neither substance would be in excess. However, if there are 3 moles of sodium and 2 moles of water, sodium is in excess and therefore water is the limiting reactant. If the chemical equation called for twice as much sodium as water, the ratio between sodium and water would be 2:1 rather than 1:1. In this case, if there were 3 moles of sodium and 2 moles of water, water would be in excess since you'd have enough water to react with 4 moles of sodium. The amount of sodium would be holding back the reaction and therefore sodium would be the limiting reactant. Hopefully, this simple example helps.

 Posts: 97
 Joined: Sat Aug 24, 2019 12:17 am
Re: Steps to calculate limiting reagant
Once you have the moles available of each reactant, you must compare to the molar ratios given in the chemical equation. Take the balanced equation: N_{2} + 3H_{2} = 2NH_{3}. You have already calculated moles of each reactant as 3.0 moles of N_{2} and 8.0 moles of H_{2}.
I would proceed by picking one of the reactants, and using the molar ratio, calculating how much of the other reactant the reaction would use if all of the first reactant was consumed. For example, if I were to start with N_{2}, I know that for every mole of N_{2} consumed, 3 moles of H_{2} are consumed. Therefore, if all 3.0 moles were consumed, the reaction would need 9.0 moles of H_{2}. Since we already know that we only have 8.0 moles of H_{2} available, H_{2} is the limiting reactant.
Note that you could work this problem in reverse by starting with 8.0 moles of H_{2}, which would need 2.67 moles of N_{2}, which is less than the 3.0 moles we have. Therefore, N_{2} is the excess reactant.
I would proceed by picking one of the reactants, and using the molar ratio, calculating how much of the other reactant the reaction would use if all of the first reactant was consumed. For example, if I were to start with N_{2}, I know that for every mole of N_{2} consumed, 3 moles of H_{2} are consumed. Therefore, if all 3.0 moles were consumed, the reaction would need 9.0 moles of H_{2}. Since we already know that we only have 8.0 moles of H_{2} available, H_{2} is the limiting reactant.
Note that you could work this problem in reverse by starting with 8.0 moles of H_{2}, which would need 2.67 moles of N_{2}, which is less than the 3.0 moles we have. Therefore, N_{2} is the excess reactant.

 Posts: 113
 Joined: Sat Jul 20, 2019 12:15 am
Re: Steps to calculate limiting reagant
Of course. And nice job acknowledging the fact that you need to use moles in chemical calculations. If given grams, the grams values must be converted to mole values through stochiometry. And this is where you left off.
I will use an example problem that could really help you out.
If we are given the balanced chemical equation: 2Al(s) + 3Cl2(s) > 2AlCl3(s) and the provided sample masses, then we can solve for the limiting reactant. In this reaction the sample masses are the following: 255. g Al and 535. g Cl2.
After mole conversations, the mole value for Al is 9.45 mol Al. Likewise, the mole value for Cl2 is 7.54 mol Cl2.
With this information, you will need to use a mole ratio operation using the provided stoichiometric coefficients from the provided balanced equation. Depending on the values of the data, an answer could either tell you that a specific substance is in excess in ratio to the other substance or limiting based on the ratio to the other substance.
When I solved this problem, I chose to determine a result using 9.45 mol Al as the reference. What I did was multiply 9.45 mol Al by the mole ratio shared between Al and Cl2, which is 2 mol Al and 3 mol Cl2. The operation was 9.45 mol Al x (3 mol Cl2/2 mol Al), and this equaled to be 14.175 mol Cl2. This value represents the required number of moles of Cl2 that is required if 9.45 mole of Al were to be reacted with it in reaction.
In essence, 9.45 mol Al requires 14.175 mol Cl2 to react with it in this specific reaction.
So with 14.175 mol Cl2 required, we can compare this to the available 7.54 mol Cl2 provided in the problem. With this data, we can conclude that Cl2 is the limiting reactant because this reaction sample containing 7.54 mol Cl2 is not enough to supply the required 14.175 mol Cl2. By identifying that Cl2 is the limiting reactant, we can then follow a normal stoichiometric operation to find the amount of AlCl2, asked in the question, that could be produced if 7.54 mol Cl2 were to be provided.
What I did was 7.54 mol Cl2 x (2 mol AlCl3/3 mol Cl2) x (133.33 g AlCl3 / 1 mole) = 669.316 g AlCl3.
If you were to find the actual percent yield, you would divide the total (Theoretical yield) by the actual yield provided in the problem, which is 300. g collected, and multiply that value by 100%.
At the end of the question, the percent yield was calculated to be 44.8% percent yield.
I hope I helped!!!
I will use an example problem that could really help you out.
If we are given the balanced chemical equation: 2Al(s) + 3Cl2(s) > 2AlCl3(s) and the provided sample masses, then we can solve for the limiting reactant. In this reaction the sample masses are the following: 255. g Al and 535. g Cl2.
After mole conversations, the mole value for Al is 9.45 mol Al. Likewise, the mole value for Cl2 is 7.54 mol Cl2.
With this information, you will need to use a mole ratio operation using the provided stoichiometric coefficients from the provided balanced equation. Depending on the values of the data, an answer could either tell you that a specific substance is in excess in ratio to the other substance or limiting based on the ratio to the other substance.
When I solved this problem, I chose to determine a result using 9.45 mol Al as the reference. What I did was multiply 9.45 mol Al by the mole ratio shared between Al and Cl2, which is 2 mol Al and 3 mol Cl2. The operation was 9.45 mol Al x (3 mol Cl2/2 mol Al), and this equaled to be 14.175 mol Cl2. This value represents the required number of moles of Cl2 that is required if 9.45 mole of Al were to be reacted with it in reaction.
In essence, 9.45 mol Al requires 14.175 mol Cl2 to react with it in this specific reaction.
So with 14.175 mol Cl2 required, we can compare this to the available 7.54 mol Cl2 provided in the problem. With this data, we can conclude that Cl2 is the limiting reactant because this reaction sample containing 7.54 mol Cl2 is not enough to supply the required 14.175 mol Cl2. By identifying that Cl2 is the limiting reactant, we can then follow a normal stoichiometric operation to find the amount of AlCl2, asked in the question, that could be produced if 7.54 mol Cl2 were to be provided.
What I did was 7.54 mol Cl2 x (2 mol AlCl3/3 mol Cl2) x (133.33 g AlCl3 / 1 mole) = 669.316 g AlCl3.
If you were to find the actual percent yield, you would divide the total (Theoretical yield) by the actual yield provided in the problem, which is 300. g collected, and multiply that value by 100%.
At the end of the question, the percent yield was calculated to be 44.8% percent yield.
I hope I helped!!!
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