Finding the Limiting Reagent
To find the limiting reagent there are a few simple steps!
Let's use the following practice problem to go through these steps so that you can see how everything should be done:
For the balanced equation shown below, what would be the limiting reagent if 27.5 grams of C6H6O were reacted with 101 grams of O2?
2C6H6O+7O2 --> 6CO2+3H2O
STEP 1:
Calculate the molar mass of each reactant (C6H6O and O2).
To do this you have to look at the periodic table (which can be found here) to find the atomic mass of each element. You will find that Carbon has an atomic mass of 12, Hydrogen has an atomic mass of 1, and Oxygen has an atomic mass of16 (as shown in the pictures below):
Let's use the following practice problem to go through these steps so that you can see how everything should be done:
For the balanced equation shown below, what would be the limiting reagent if 27.5 grams of C6H6O were reacted with 101 grams of O2?
2C6H6O+7O2 --> 6CO2+3H2O
STEP 1:
Calculate the molar mass of each reactant (C6H6O and O2).
To do this you have to look at the periodic table (which can be found here) to find the atomic mass of each element. You will find that Carbon has an atomic mass of 12, Hydrogen has an atomic mass of 1, and Oxygen has an atomic mass of16 (as shown in the pictures below):
Since you have found the atomic mass, you multiply each atomic mass by the amount of atoms each element has in the reactant to find the molar mass. We know that there are 6 atoms of Carbon, 6 atoms of Hydrogen, and only one atom of Oxygen in the first reactant since it was given to us in the original equation--> C6H6O see?!
It's simple, multiply:
It's simple, multiply:
Likewise, we calculate the molar mass of O2 :
STEP 2:
Calculate the number of moles for each reactant.
To do this you would use the following equation and plug in the quantities that were given to you in the original question:
Calculate the number of moles for each reactant.
To do this you would use the following equation and plug in the quantities that were given to you in the original question:
Since we know the # of Grams (highlighted in the original equation below) we can plug in all of our known quantities to calculate the number of moles.
Using the formula for finding the # of Moles, we use the following steps to find that C6H6O produces .29 moles and O2 produces 3.12 moles :
STEP 3:
Determine the mole ratio of the reactants.
We finally look at the actual equation that was given to us in the problem (2C6H6O+7O2 --> 6CO2+3H2O) however, we only want to use the part of the equation that is before the arrow (2C6H6O+7O2). To find the mole ratio, we find the coefficient of each reactant (the large number that is in front of each reactant).
Determine the mole ratio of the reactants.
We finally look at the actual equation that was given to us in the problem (2C6H6O+7O2 --> 6CO2+3H2O) however, we only want to use the part of the equation that is before the arrow (2C6H6O+7O2). To find the mole ratio, we find the coefficient of each reactant (the large number that is in front of each reactant).
STEP 4:
Set up a proportion and cross multiply.
The proportion that we set up will cross multiply the number of moles each reactant has (found in step 2) by the mole ratio of the reactants (found in step 3). As always, look below for a drawn explanation:
Set up a proportion and cross multiply.
The proportion that we set up will cross multiply the number of moles each reactant has (found in step 2) by the mole ratio of the reactants (found in step 3). As always, look below for a drawn explanation:
You multiply the number of moles of one reactant with the other reactant's coefficient and vise versa.
When calculated you will find that .29*7= 2.03 (this is for C6H6O) and 3.12*2=6.24 (this is for O2).
Since 2.03 is smaller than 6.24, C6H6O is the limiting reagent since it will get used up sooner.
Now that you have read this lesson on how to find limiting reagents, click here for some practice problems
