Limiting Reagent
What the heck is a limiting reagent?
Imagine you're baking a cake. A very simple cake whose recipe calls for 2 eggs and 2 cups of flour. At your house, you have 4 eggs and 10 cups of flour. How many cakes can you bake (without going to the store to buy more baking supplies)?
Most likely, you were able to (correctly) calculate that you can bake 2 cakes.
How did you figure this out? It's probable that you did the calculations in your head so quickly, you don't even know how you got your answer. If that's true, observe the following:
4 eggs x 1 cake = 2 cakes
2 eggs
10 c flour x 1 cakes = 5 cakes
2 c flour
The dimensional analysis calculations above show that the amount of eggs you have (4) will allow you to make 2 cakes, while the amount of cups of flour you have (10) will allow you to bake 5 cakes. Even though you have enough flour to bake 5 cakes, you can only make 2. Why? You only have enough eggs to make 2 cakes, so it really doesn't matter how many cakes you can make with the amount of flour you have.
The number of eggs you have is the limiting reagent. After you bake your 2 cakes, you will have flour in excess (left over).
So, in any chemical reaction, the limiting reagent is the reactant that limits the amount of product that can be formed. The excess reagent is the reactant that remains after the chemical reaction takes place.
How can I find the limiting reagent?
In a chemical reaction, there are multiple ways to find the limiting reagent. You will now be shown 3 of them. Feel free to use whichever way you understand best.
#1
This method is exactly the one shown above in our cake example. This method entails using the amount of each reactant you have to determine how much product each will produce. The limiting reagent is the reactant that yields less product. The other reactant is in excess.
To do this, we must use molar masses, and mole ratios (observed in the balanced chemical equation).
For example: You have 10 grams of Na and 10 grams of H20. Reacting these two specimen together can be shown by the following chemical equation: 2Na + 2H20 --> 2NaOH + H2 What is the maximum number of grams of H2 that can be produced?
10 g Na x 1 mol Na x 1 mol H2 x 2 g H2 = 0.435 g H2
23 g Na 2 mol Na 1 mol H2
10 g H20 x 1 mol H2O x 1 mol H2 x 2 g H2 = 0.556 g H2
18 g H2O 2 mol H2O 1 mol H2
When all 10 grams of Na react, 0.435 grams of H2 are produced. When all 10 grams of H2O react, 0.556 g of H2 are produced. This shows us that Na is the limiting reagent, and that H2O is the excess reagent.
A maximum of 0.435 g of H2 can be produced.
Tip:
Molar masses are found by calculating the mass of each mole of a substance. The molar mass for each element can be found on a Periodic Table of Elements.
Molar ratios can be observed from the balanced chemical equation. For example, the equation above tells us that for every 1 mole of H2 produced, 2 moles of Na must react.
#2
This method of finding the limiting and excess reagents uses dimensional analysis to go from grams of the first reactant, to grams of the second reactant. This tells us how much of reactant 1 will be needed to react with a certain amount of reactant 2 (and vice versa).
For example: If you start with 25 grams of lead(II)nitrate and 15 grams of sodium iodide, how many grams of sodium nitrate can be formed?
Lead(II)nitrate and sodium iodide are both reactants. Sodium nitrate is a product.
Our first step is to determine which of our 2 reactants is the limiting reagent. We will do this by seeing how many grams of sodium iodide are required to react completely with the 25 grams of lead(II)nitrate, using dimensional analysis.
25 g Pb(NO3)2 x 1 mol Pb(NO3)2 x 2 mol NaI x 149.9 g NaI = 21.7 g NaI
345 g Pb(NO3)2 1 mol Pb(NO3)2 1 mol NaI
We now know that it will take 21.7 grams of sodium iodide to react completely with 25 grams of lead(II)nitrate. However, we only have 15 grams of sodium iodide to react. This tells us that sodium iodide is our limiting reagent, and that we will have lead(II)nitrate in excess.
Now we want to know how much sodium nitrate will be produced during this reaction. We will use dimensional analysis again, this time starting with our limiting reagent. It doesn't matter how many grams of sodium nitrate can be produced by 25 grams of lead(II)nitrate, because we do not have enough sodium iodide to complete this reaction (much like we did not have enough eggs to make 5 cakes!).
15 g NaI x 1 mol NaI x 2 mol NaNO3 x 85 g NaNO3 = 8.51 g NaNO3
149.9 g NaI 1 mol NaI 1 mol NaNO3
Taking the limiting reagent into account, 8.51g of sodium nitrate can be formed.
#3
This method requires us to determine the moles of reactant 1 and the moles of reactant 2. We must then use these 2 values to form a ratio, and determine which is the limiting reagent, and which is the excess reagent, once again using dimensional analysis.
For example: You have 14.8 grams of C3H8 and 3.44 grams of O2. You also have the following chemical equation: C3H8 + 5O2 --> 3CO2 + 4H2O Which reactant would be the limiting reagent during this chemical reaction?
First, we must turn 14.8 g of C3H8 into moles of C3H8. This is done using a molar mass.
14.8 g C3H8 x 1 mol C3H8 = 0.336 mol C3H8
44 g C3H8
We must then do a similar calculation to find how many moles of O2 makes 3.44 grams of O2.
3.44 g O2 x 1 mol O2 = 0.108 mol O2
32 g O2
Next, we divide these 2 values to determine the ratio of reactants.
0.336 mol C3H8 = 3.11 mol C3H8
0.108 mol 02 1 mol O2
We now compare this to the true combining mole ratio (found in the balanced chemical equation).
true combining mole ratio = 1 mol C3H8
5 mol O2
This comparison tells us that for every 3.11 moles of C3H8 we would need (3.11 x 5 = approx. 15) approximately 15 moles of O2. We only have 1 mole of O2. This means that O2 is the limiting reagent, and that we will have excess C3H8.
Extra Stuff
Videos
Limiting Reagent Video (Step by step explanation video)
Another Video (Similar to the first, but with a slightly different explanation)
Worksheets
Use any of these worksheets to test yourself!
Worksheet #1 (Contains answers)
Worksheets 2 & 3 (Contains answers)
Worksheet 4 (Contains answers)
A Great Website
Chem Team (Check it out!)
Most likely, you were able to (correctly) calculate that you can bake 2 cakes.
How did you figure this out? It's probable that you did the calculations in your head so quickly, you don't even know how you got your answer. If that's true, observe the following:
4 eggs x 1 cake = 2 cakes
2 eggs
10 c flour x 1 cakes = 5 cakes
2 c flour
The dimensional analysis calculations above show that the amount of eggs you have (4) will allow you to make 2 cakes, while the amount of cups of flour you have (10) will allow you to bake 5 cakes. Even though you have enough flour to bake 5 cakes, you can only make 2. Why? You only have enough eggs to make 2 cakes, so it really doesn't matter how many cakes you can make with the amount of flour you have.
The number of eggs you have is the limiting reagent. After you bake your 2 cakes, you will have flour in excess (left over).
So, in any chemical reaction, the limiting reagent is the reactant that limits the amount of product that can be formed. The excess reagent is the reactant that remains after the chemical reaction takes place.
How can I find the limiting reagent?
In a chemical reaction, there are multiple ways to find the limiting reagent. You will now be shown 3 of them. Feel free to use whichever way you understand best.
#1
This method is exactly the one shown above in our cake example. This method entails using the amount of each reactant you have to determine how much product each will produce. The limiting reagent is the reactant that yields less product. The other reactant is in excess.
To do this, we must use molar masses, and mole ratios (observed in the balanced chemical equation).
For example: You have 10 grams of Na and 10 grams of H20. Reacting these two specimen together can be shown by the following chemical equation: 2Na + 2H20 --> 2NaOH + H2 What is the maximum number of grams of H2 that can be produced?
10 g Na x 1 mol Na x 1 mol H2 x 2 g H2 = 0.435 g H2
23 g Na 2 mol Na 1 mol H2
10 g H20 x 1 mol H2O x 1 mol H2 x 2 g H2 = 0.556 g H2
18 g H2O 2 mol H2O 1 mol H2
When all 10 grams of Na react, 0.435 grams of H2 are produced. When all 10 grams of H2O react, 0.556 g of H2 are produced. This shows us that Na is the limiting reagent, and that H2O is the excess reagent.
A maximum of 0.435 g of H2 can be produced.
Tip:
Molar masses are found by calculating the mass of each mole of a substance. The molar mass for each element can be found on a Periodic Table of Elements.
Molar ratios can be observed from the balanced chemical equation. For example, the equation above tells us that for every 1 mole of H2 produced, 2 moles of Na must react.
#2
This method of finding the limiting and excess reagents uses dimensional analysis to go from grams of the first reactant, to grams of the second reactant. This tells us how much of reactant 1 will be needed to react with a certain amount of reactant 2 (and vice versa).
For example: If you start with 25 grams of lead(II)nitrate and 15 grams of sodium iodide, how many grams of sodium nitrate can be formed?
Lead(II)nitrate and sodium iodide are both reactants. Sodium nitrate is a product.
Our first step is to determine which of our 2 reactants is the limiting reagent. We will do this by seeing how many grams of sodium iodide are required to react completely with the 25 grams of lead(II)nitrate, using dimensional analysis.
25 g Pb(NO3)2 x 1 mol Pb(NO3)2 x 2 mol NaI x 149.9 g NaI = 21.7 g NaI
345 g Pb(NO3)2 1 mol Pb(NO3)2 1 mol NaI
We now know that it will take 21.7 grams of sodium iodide to react completely with 25 grams of lead(II)nitrate. However, we only have 15 grams of sodium iodide to react. This tells us that sodium iodide is our limiting reagent, and that we will have lead(II)nitrate in excess.
Now we want to know how much sodium nitrate will be produced during this reaction. We will use dimensional analysis again, this time starting with our limiting reagent. It doesn't matter how many grams of sodium nitrate can be produced by 25 grams of lead(II)nitrate, because we do not have enough sodium iodide to complete this reaction (much like we did not have enough eggs to make 5 cakes!).
15 g NaI x 1 mol NaI x 2 mol NaNO3 x 85 g NaNO3 = 8.51 g NaNO3
149.9 g NaI 1 mol NaI 1 mol NaNO3
Taking the limiting reagent into account, 8.51g of sodium nitrate can be formed.
#3
This method requires us to determine the moles of reactant 1 and the moles of reactant 2. We must then use these 2 values to form a ratio, and determine which is the limiting reagent, and which is the excess reagent, once again using dimensional analysis.
For example: You have 14.8 grams of C3H8 and 3.44 grams of O2. You also have the following chemical equation: C3H8 + 5O2 --> 3CO2 + 4H2O Which reactant would be the limiting reagent during this chemical reaction?
First, we must turn 14.8 g of C3H8 into moles of C3H8. This is done using a molar mass.
14.8 g C3H8 x 1 mol C3H8 = 0.336 mol C3H8
44 g C3H8
We must then do a similar calculation to find how many moles of O2 makes 3.44 grams of O2.
3.44 g O2 x 1 mol O2 = 0.108 mol O2
32 g O2
Next, we divide these 2 values to determine the ratio of reactants.
0.336 mol C3H8 = 3.11 mol C3H8
0.108 mol 02 1 mol O2
We now compare this to the true combining mole ratio (found in the balanced chemical equation).
true combining mole ratio = 1 mol C3H8
5 mol O2
This comparison tells us that for every 3.11 moles of C3H8 we would need (3.11 x 5 = approx. 15) approximately 15 moles of O2. We only have 1 mole of O2. This means that O2 is the limiting reagent, and that we will have excess C3H8.
Extra Stuff
Videos
Limiting Reagent Video (Step by step explanation video)
Another Video (Similar to the first, but with a slightly different explanation)
Worksheets
Use any of these worksheets to test yourself!
Worksheet #1 (Contains answers)
Worksheets 2 & 3 (Contains answers)
Worksheet 4 (Contains answers)
A Great Website
Chem Team (Check it out!)