Limiting Reagents
When there is an insufficient amount of a certain reactant in a chemical equation, the reaction ceases to continue. By finding which reactant will, “limit,” the chemical reaction and which reactant will be in, “excess,” one can determine the amount of product that was produced. Excess means that more than enough of a certain compound will be produced, therefore another compound will run out first. The limiting reagent is defined simply as the substance in a chemical reaction that runs out first.
You could also define the limiting reagent as the reactant that is completely used up in a reaction and thus determines when the reaction stops. From stoichiometry one can calculate the exact amount of reactant needed to react with another element.
Example:
"After 20 g. of A and 20 g. of B react, how much of the excess compound remains?" To answer this problem, subtract the limiting reagent amount from the excess amount.
An actual limiting reagent problem you encounter may look like this:
Example: A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl according to the reaction:
Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O
What is the theoretical yield of MgCl2?
A strategy to ensure you know how to properly approach a limiting reagent problem such as this is to write all the numbers directly below the equation.
Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O
50.6 g 45.0g ? g
In the next step to solving the problem, covert each of the grams in the reactants to moles. This is done by multiplying the amount of grams given by 1 over the molar mass of the compound.
Choose either one of these reactants and calculate how many moles of the other reactant is needed to completely use up the reactant chosen. If you were to try using magnesium hydroxide:
Compare the moles of HCl needed to the actual moles of HCl available. In this case, 1.74 mol of HCl is needed and 1.23 mol HCl is available, which is not enough. So, although it appears that there are more moles of HCl present than Mg(OH)2, the HCl is the limiting reagent. The HCl will be run out before the magnesium hydroxide and thereby limit the amount of product formed.
It is common, however, that one will also be asked to to calculate, “theoretical yield,” and, “percent yield,” after identifying the limiting reagent. The theoretical yield is the ideal, maximum amount of product that could theoretically be produced. Percent yield is used as a sort of measuring device to show how close to the most ideal result one has obtained in a chemical synthesis. By taking the moles of HCl and using dimensional analysis, multiplying moles and the molar mass of MgCl2, the theoretical yield is obtained.
Suppose in the same reaction it was told to you that a chemist actually obtained 55.4 g of MgCl2. This could be called the actual yield. Dividing the actual yield by the theoretical yield would grant you the desired percent yield for the problem.
Consider viewing a second typical limiting reagent problem one might run across.
Example Problem 2
2 Al + 3 I2 ------> 2 AlI3
Determine the limiting reagent and the theoretical yield of the product if one starts with:
a) 1.20 mol Al and 2.40 mol iodine.
b) 1.20 g Al and 2.40 g iodine
c) How many grams of Al are left over in part b?
To determine the limiting reagent: take the given moles of each substance and divide it by the coefficient of the balanced equation. The substance with the smallest amount as an answer is the limiting reagent.
This technique is essential and will come up time and time again, so it is worthwhile to remember.
For aluminum: 1.20 / 2 = 0.60
For iodine: 2.40 / 3 = 0.80
The lowest number is 0.60 for aluminum, making it the limiting reagent. Aluminum will run out first in part a.
1.20/2 means there are 0.60 "groupings" of 2 and 2.40/3 means there are 0.80 "groupings" of 3. If they ran out at the same time, we'd need one "grouping" of each. Since there is less of the "grouping of 2," it will run out first.
The second part of the question, "theoretical yield," is dependent on determining the limiting reagent, as was shown in the previous example problem. Once that is done, it becomes a stoichiometric calculation. Stoichiometric calculations are further described in the Stoichiometry section of our website.
Al and AlI3 stand in a 1 to 1 molar relationship, so 1.20 mol of Al produces 1.20 mol of AlI3. Notice that the amount of I2 does not play a role because it’s in excess
Part B Solution: since grams are given, the grams must be converted to moles.
For the mole calculation:
aluminum is 1.20 g / 26.98 g mol¯1 = 0.04477 mol
iodine is 2.4 g / 253.8 g mol¯1 = 0.009456 mol
To determine the limiting reagent:
aluminum is 0.04477 / 2 = 0.02238
iodine is 0.009456 / 3 = 0.003152
The lower number is iodine, that is identified as the limiting reagent.
Finally, you have to perform a calculation that will involve the iodine, but not the aluminum.
I2 and AlI3 stand in a 3 to 2 molar relationship, so 0.009456 mol of I2 produces 0.006304 mol of AlI3. Again, notice that the amount of Al does not play a role, since it is in excess.
From here figure out the grams of AlI3 and that is the final answer.
Part C Solution: Since the moles were calculated already, you can calculate directly and then convert to grams.
Al and I2 stand in a 2 to3 molar relationship, so 0.009456 mol of I2 uses 0.006304 mol of Al.
Convert this aluminum amount to grams and subtract it from 1.20 g and you get the answer.
Review
A brief summation of techniques one can use to find the limiting reagent are displayed here:
Formula 1: Find the limiting reagent by looking at the number of moles of each reactant.
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + energy
What mass of carbon dioxide forms in the reaction of 25 grams of glucose with 40 grams of oxygen?
When approaching this problem, you should be able to see that for every 1 mole of glucose (C6H12O6), you need 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction. This is already given.
Step 2: Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).
25g x (1 mol/180.06g) = 0.1388 mol C6H12O6
40g x (1 mol/32g) = 1.25 mol O2
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
a. If all of the 1.25 moles of oxygen were to be used up, there would need to be 1.25 x (1/6) or 0.208 moles of glucose. There is only 0.1388 moles of glucose available which makes it the limiting reactant.
1.25 mol O2 x (1 mol C6H12O6)/(6 mol O2) = 0.208 mol C6H12O6
b. If all of the 0.1388 moles of glucose were to be used up, there would need to be 0.1388 x 6 or 0.8328 moles of oxygen. Since there is an excess of oxygen, the glucose amount will be used to calculate the amount of the products in the reaction.
0.1388 mol C6H12O6 x (6 mol O2)/(1 mol C6H12O6) = 0.8328 mol O2
If more than 6 moles of O2 is available per mole of C6H12O6, the oxygen is in excess and the glucose is the limiting reactant. If less than 6 mole of oxygen is available per mole of glucose, the oxygen is the limiting reactant. Our ratio is 6 mole oxygen per 1 mole glucose, OR 1 mole oxygen per 1/6 mole glucose. This means: 6 mol O2 / 1 mol C6H12O6 .
Therefore the mole ratio is: (0.8328 mol O2)/(0.208 mol C6H12O6)
Which gives us a 4.004 ratio of O2 to C6H12O6.
Step 4: Use the amount of limiting reactant to calculate the amount of CO2 or H2O produced.
For carbon dioxide produced: 0.1388 moles glucose x (6/1) = 0.8328 moles carbon dioxide.
Step 5: If necessary, calculate how much is left in excess.
1.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over
Formula 2: Find the limiting reagent by calculating and comparing the amount of product each reactant will produce.
Calculate the mass of magnesium oxide possible if 2.40g Mg reacts with 10.0g O2
Step 1: Balance equation
2 Mg + O2 → 2 MgO
Step 2 and Step 3: Converting mass to moles and stoichiometry
(2.40g Mg) x (1.00 mol Mg)/(24.31g Mg) x (2.00 mol MgO)/(2.00 mol Mg) x (40.31g MgO)/(1.00 mol MgO) = 3.98g MgO
(10.0g O2) x (1 mol O2)/(32.0g O2) x (2 mol MgO)/(1 mol O2) x (40.31g MgO)/(1 mol MgO) = 25.2g MgO
Step 4: The reactant that produces a lesser amount of product is the limiting reagent
Mg produces less amount of MgO than O2 (3.98g MgO vs. 25.2g MgO), therefore Mg is the limiting reagent in this reaction.
Step 5: The reactant that produces a higher amount of product is the excess reagent
O2 produces more amount of MgO than Mg (25.2g MgO vs. 3.98 MgO), therefore O2 is the excess reagent in this reaction.
Step 6: Find the amount of remaining excess reactant by subtracting the mass of the excess reagent consumed from the total mass of excess
reagent given
Mass of excess reagent calculated using the limiting reagent:
(2.40g Mg) x (1.00 mol Mg)/(24.31g Mg) x (1.00 mol O2)/(2.00 mol Mg) x (32.0g O2)/(1.00 mol O2) = 1.58g O2
OR Mass of excess reagent calculated using the mass of the product:
(3.98g MgO) x (1.00 mol MgO)/(40.31g MgO) x (1.00 mol O2)/(2.00 mol MgO) x (32.0g O2)/(1.00 mol O2) = 1.58g O2
Mass of total excess reagent given – mass of excess reagent consumed in the reaction
10.0g – 1.58g = 8.42g O2 is in excess.
You could also define the limiting reagent as the reactant that is completely used up in a reaction and thus determines when the reaction stops. From stoichiometry one can calculate the exact amount of reactant needed to react with another element.
Example:
"After 20 g. of A and 20 g. of B react, how much of the excess compound remains?" To answer this problem, subtract the limiting reagent amount from the excess amount.
An actual limiting reagent problem you encounter may look like this:
Example: A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl according to the reaction:
Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O
What is the theoretical yield of MgCl2?
A strategy to ensure you know how to properly approach a limiting reagent problem such as this is to write all the numbers directly below the equation.
Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O
50.6 g 45.0g ? g
In the next step to solving the problem, covert each of the grams in the reactants to moles. This is done by multiplying the amount of grams given by 1 over the molar mass of the compound.
Choose either one of these reactants and calculate how many moles of the other reactant is needed to completely use up the reactant chosen. If you were to try using magnesium hydroxide:
Compare the moles of HCl needed to the actual moles of HCl available. In this case, 1.74 mol of HCl is needed and 1.23 mol HCl is available, which is not enough. So, although it appears that there are more moles of HCl present than Mg(OH)2, the HCl is the limiting reagent. The HCl will be run out before the magnesium hydroxide and thereby limit the amount of product formed.
It is common, however, that one will also be asked to to calculate, “theoretical yield,” and, “percent yield,” after identifying the limiting reagent. The theoretical yield is the ideal, maximum amount of product that could theoretically be produced. Percent yield is used as a sort of measuring device to show how close to the most ideal result one has obtained in a chemical synthesis. By taking the moles of HCl and using dimensional analysis, multiplying moles and the molar mass of MgCl2, the theoretical yield is obtained.
Suppose in the same reaction it was told to you that a chemist actually obtained 55.4 g of MgCl2. This could be called the actual yield. Dividing the actual yield by the theoretical yield would grant you the desired percent yield for the problem.
Consider viewing a second typical limiting reagent problem one might run across.
Example Problem 2
2 Al + 3 I2 ------> 2 AlI3
Determine the limiting reagent and the theoretical yield of the product if one starts with:
a) 1.20 mol Al and 2.40 mol iodine.
b) 1.20 g Al and 2.40 g iodine
c) How many grams of Al are left over in part b?
To determine the limiting reagent: take the given moles of each substance and divide it by the coefficient of the balanced equation. The substance with the smallest amount as an answer is the limiting reagent.
This technique is essential and will come up time and time again, so it is worthwhile to remember.
For aluminum: 1.20 / 2 = 0.60
For iodine: 2.40 / 3 = 0.80
The lowest number is 0.60 for aluminum, making it the limiting reagent. Aluminum will run out first in part a.
1.20/2 means there are 0.60 "groupings" of 2 and 2.40/3 means there are 0.80 "groupings" of 3. If they ran out at the same time, we'd need one "grouping" of each. Since there is less of the "grouping of 2," it will run out first.
The second part of the question, "theoretical yield," is dependent on determining the limiting reagent, as was shown in the previous example problem. Once that is done, it becomes a stoichiometric calculation. Stoichiometric calculations are further described in the Stoichiometry section of our website.
Al and AlI3 stand in a 1 to 1 molar relationship, so 1.20 mol of Al produces 1.20 mol of AlI3. Notice that the amount of I2 does not play a role because it’s in excess
Part B Solution: since grams are given, the grams must be converted to moles.
For the mole calculation:
aluminum is 1.20 g / 26.98 g mol¯1 = 0.04477 mol
iodine is 2.4 g / 253.8 g mol¯1 = 0.009456 mol
To determine the limiting reagent:
aluminum is 0.04477 / 2 = 0.02238
iodine is 0.009456 / 3 = 0.003152
The lower number is iodine, that is identified as the limiting reagent.
Finally, you have to perform a calculation that will involve the iodine, but not the aluminum.
I2 and AlI3 stand in a 3 to 2 molar relationship, so 0.009456 mol of I2 produces 0.006304 mol of AlI3. Again, notice that the amount of Al does not play a role, since it is in excess.
From here figure out the grams of AlI3 and that is the final answer.
Part C Solution: Since the moles were calculated already, you can calculate directly and then convert to grams.
Al and I2 stand in a 2 to3 molar relationship, so 0.009456 mol of I2 uses 0.006304 mol of Al.
Convert this aluminum amount to grams and subtract it from 1.20 g and you get the answer.
Review
A brief summation of techniques one can use to find the limiting reagent are displayed here:
Formula 1: Find the limiting reagent by looking at the number of moles of each reactant.
- Determine the balanced chemical equation for the chemical reaction.
- Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).
- Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
- Use the amount of limiting reactant to calculate the amount of product produced.
- If necessary, calculate how much is left in excess of the non-limiting reagent.
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + energy
What mass of carbon dioxide forms in the reaction of 25 grams of glucose with 40 grams of oxygen?
When approaching this problem, you should be able to see that for every 1 mole of glucose (C6H12O6), you need 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction. This is already given.
Step 2: Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).
25g x (1 mol/180.06g) = 0.1388 mol C6H12O6
40g x (1 mol/32g) = 1.25 mol O2
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
a. If all of the 1.25 moles of oxygen were to be used up, there would need to be 1.25 x (1/6) or 0.208 moles of glucose. There is only 0.1388 moles of glucose available which makes it the limiting reactant.
1.25 mol O2 x (1 mol C6H12O6)/(6 mol O2) = 0.208 mol C6H12O6
b. If all of the 0.1388 moles of glucose were to be used up, there would need to be 0.1388 x 6 or 0.8328 moles of oxygen. Since there is an excess of oxygen, the glucose amount will be used to calculate the amount of the products in the reaction.
0.1388 mol C6H12O6 x (6 mol O2)/(1 mol C6H12O6) = 0.8328 mol O2
If more than 6 moles of O2 is available per mole of C6H12O6, the oxygen is in excess and the glucose is the limiting reactant. If less than 6 mole of oxygen is available per mole of glucose, the oxygen is the limiting reactant. Our ratio is 6 mole oxygen per 1 mole glucose, OR 1 mole oxygen per 1/6 mole glucose. This means: 6 mol O2 / 1 mol C6H12O6 .
Therefore the mole ratio is: (0.8328 mol O2)/(0.208 mol C6H12O6)
Which gives us a 4.004 ratio of O2 to C6H12O6.
Step 4: Use the amount of limiting reactant to calculate the amount of CO2 or H2O produced.
For carbon dioxide produced: 0.1388 moles glucose x (6/1) = 0.8328 moles carbon dioxide.
Step 5: If necessary, calculate how much is left in excess.
1.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over
Formula 2: Find the limiting reagent by calculating and comparing the amount of product each reactant will produce.
- Balance the chemical equation for the chemical reaction.
- Convert the given information into moles.
- Use stoichiometry for each individual reactant to find the mass of product produced.
- The reactant that produces a lesser amount of product is the limiting reagent.
- The reactant that produces a higher amount of product is the excess reagent.
- To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
Calculate the mass of magnesium oxide possible if 2.40g Mg reacts with 10.0g O2
Step 1: Balance equation
2 Mg + O2 → 2 MgO
Step 2 and Step 3: Converting mass to moles and stoichiometry
(2.40g Mg) x (1.00 mol Mg)/(24.31g Mg) x (2.00 mol MgO)/(2.00 mol Mg) x (40.31g MgO)/(1.00 mol MgO) = 3.98g MgO
(10.0g O2) x (1 mol O2)/(32.0g O2) x (2 mol MgO)/(1 mol O2) x (40.31g MgO)/(1 mol MgO) = 25.2g MgO
Step 4: The reactant that produces a lesser amount of product is the limiting reagent
Mg produces less amount of MgO than O2 (3.98g MgO vs. 25.2g MgO), therefore Mg is the limiting reagent in this reaction.
Step 5: The reactant that produces a higher amount of product is the excess reagent
O2 produces more amount of MgO than Mg (25.2g MgO vs. 3.98 MgO), therefore O2 is the excess reagent in this reaction.
Step 6: Find the amount of remaining excess reactant by subtracting the mass of the excess reagent consumed from the total mass of excess
reagent given
Mass of excess reagent calculated using the limiting reagent:
(2.40g Mg) x (1.00 mol Mg)/(24.31g Mg) x (1.00 mol O2)/(2.00 mol Mg) x (32.0g O2)/(1.00 mol O2) = 1.58g O2
OR Mass of excess reagent calculated using the mass of the product:
(3.98g MgO) x (1.00 mol MgO)/(40.31g MgO) x (1.00 mol O2)/(2.00 mol MgO) x (32.0g O2)/(1.00 mol O2) = 1.58g O2
Mass of total excess reagent given – mass of excess reagent consumed in the reaction
10.0g – 1.58g = 8.42g O2 is in excess.
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