Welcome to Limiting Reagent Made Easy brought to you by Ketzbook
This is the final tutorial in my stoichiometry series, and in this video, we will learn how
to identify reactants as limiting or in excess and how to calculate reaction yields in a
limiting reagent problem.
The limiting reagent (aka limiting reactant) of a reaction is the reactant that limits
the amount product formed.
Let’s see how this works with s’mores.
Suppose you went camping and took 1 box of graham crackers (which contains 27 crackers),
1 bag of marshmallows (which contains 40 marshmallows), and 12 bars of chocolate. How many s’mores
could you make if you followed the original s’mores recipe?
The original s’mores recipe, which first appeared in the 1927 Girl Scout Handbook and
calls the dessert “some more,” states that you need 2 marshmallows toasted to a
crisp gooey state, one graham cracker split in half, and a half of a bar of chocolate
to make one s’more. In order to get rid of the one half, let’s just double the recipe,
so 4 marshmallows, 2 graham crackers, and one chocolate bar are needed to make 2 s’mores.
Now, in order to solve this problem, we need to calculate the number of s’mores that
each reactant could make.
Start with the graham crackers. You have 27 graham crackers, and 2 graham crackers can
make 2 s’mores, so you could make up to 27 s’mores from the amount of graham crackers you have.
You have 40 marshmallows, and it takes 4 marshmallows to make 2 s’mores, so you could make up
to 20 s’mores from the amount of marshmallows you have.
You have 12 bars of chocolate, and one bar of chocolate can make 2 s’mores, so you
could make up to 24 s’mores from the amount of chocolate you have.
Now, sit back and look at the different amount of s’mores that you could make from the
different reactants. What do you think? How many s’mores can you actually make?
That’s right, you can only make 20 s’mores because after that you won’t have any more marshmallows.
Marshmallows are the limiting reagent because they can make the least amount of s’mores.
The limiting reagent is the reactant that can make the least amount of product.
The other reactants are in excess.
What about the theoretical yield? How much product can we make if everything works perfectly?
That’s right, 20 s’mores. The amount of product made by the limiting reagent is the theoretical yield.
The theoretical yield is always the smallest amount of product that you calculate.
All right, now let’s try this with some chemicals. Suppose that we combined 75 grams
of magnesium with 45 grams of nitrogen and heated them up. Predict the product of the reaction,
determine the limiting reagent, and calculate the theoretical yield.
Like always, start by writing down the balanced reaction. Magnesium and nitrogen are the reactants.
Remember that nitrogen is diatomic. This is a combination reaction of a metal and a non-metal,
so there should be one product, which is an ionic compound. Magnesium will lose its two
valence electrons to become Mg2+, and nitrogen will gain three valence electrons to get a
total of 8 valence electrons and become N3-. The formula of the product, magnesium nitride,
is Mg3N2 because the numbers of the charges switch places and become subscripts.
As it is written, the reaction is not balanced, but we can fix that by writing a 3 in front of Mg,
making 3 magnesiums on both sides of the reaction.
Next, we should write down the molar masses of all the reactants and products.
Magnesium is 24.3 grams per mole, nitrogen is 28 grams per mole,
and magnesium nitride 100.9 grams per mole.
In order to determine both the limiting reactant and the theoretical yield, we need to calculate
the amount of product that EACH reactant could make. The smallest amount of product is the
theoretical yield, and the reactant that makes the smallest amount of product is the limiting reagent.
Let’s start with magnesium. We have 75 grams of magnesium, and we want to figure out how
much magnesium nitride we can make from it, so we multiply it by 3 conversion factor fractions.
As always, we write “1 mole” on the top left and the bottom right. On the bottom left
goes the mass of 1 mole of magnesium, that is 24.3 grams. Grams on the top and bottom cancel out.
In the middle fraction, we put chemicals with their coefficients from the balanced reaction.
Because we are starting with magnesium, write 3 Mg on the bottom. Because we are solving
for magnesium nitride, write 1 Mg3N2 on the top. Mg on the top and bottom cancel each other out.
Lastly, write the molar mass of magnesium nitride in the top right, 100.9 grams.
Moles on the top and bottom cancel out. Calculate from left to right, multiplying by numbers
on the top of fractions and dividing by numbers on the bottom of fractions. In your calculator,
type 75 divided by 24.3 divided by 3 times 100.9, and the answer works out to be 104
grams of magnesium nitride.
Next, we need to repeat this entire process for each reactant. So, go ahead and write
down 45 grams of nitrogen, and multiply it by 3 conversion factor fractions. Once again,
“1 mole” goes in the top left and the bottom right. Because we are solving for exactly
the same product, every numerator must be the same as our previous calculation.
This is always true for limiting reagent questions. Go ahead and copy them from your previous set up,
1 Mg3N2 in the middle, and 100.9 grams on the right.
The only two numbers that will be different are the bottom left and the bottom middle
because they correspond to the reactant. On the bottom left goes the molar mass of nitrogen,
that is 28 grams. And in the bottom middle we write “1 N2” from the balanced reaction.
To calculate this we type 45 divided by 28 times 100.9, which is 162 grams of magnesium nitride.
Now pause the video and try to answer the questions: what is the theoretical yield,
and what is the limiting reagent?
That’s right! 104 grams of magnesium nitride is the smallest amount of product, so it must
be the theoretical yield. It is impossible to make more than that because you will run
out of magnesium. The reactant that makes the smallest amount of product is the limiting reagent,
so magnesium is the limiting reagent in this case.
After the reaction is done, there will be some of the other reactant left over,
so we say that there is excess nitrogen for this reaction.
Let’s take this problem one step further.
Suppose that we actually performed this reaction in the lab and collected 88.7 grams of magnesium nitride.
Calculate the percent yield of the reaction.
Remember that the percent yield is the actual yield divided by the theoretical yield all
multiplied by 100 to turn the fraction into a percent. Also, remember that the actual yield
is always found in the question itself. How much did we make? 88.7 grams.
That is the actual yield. Let’s plug in the numbers to calculate the percent yield. 88.7 grams
divided by 104 grams times 100 percent works out to be 85%, which is a very reasonable yield.
Last of all, remember that the limiting reagent is the reactant that produces the least amount
of product, and that least amount of product is the theoretical yield.
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