When you see chemical reactions written out in textbooks, it gives you the idea that you
always have the exact right amounts for all of the reactants to be consumed completely
in the reaction.
In real life, in the lab, this is not the case.
Usually you wind up using all of one reactant and then having some of the other reactants
left over.
The reactant that is completely used up first is called the Limiting Reactant
or Limiting Reagent.
We say the other reactants are present IN EXCESS.
Here’s an example you might come across in your real life.
You want to make some Peanut Butter and Jelly sandwiches for you and your friends.
To make a PB&J, you need: 2 slices of bread
2 Tablespoons of PB 1 Tablespoon of Jelly.
If we were to write this like a chemical equation, it would look like this:
2 slices of bread + 2 T Peanut Butter + 1 T Jelly → 1 PB&J
But what if you have in the cupboard 10 slices of bread, 6 T PB, and 4 T Jelly?
Which ingredient are you going to run out of first?In this case, you can eyeball it
and quickly come to the conclusion that you’ll run out of PB first.
You have enough bread for 5 sandwiches, enough PB for 3 sandwiches, and enough Jelly for
4 sandwiches.
So in this scenario, PB is the limiting reagent.
You have excess bread and jelly.
Now let’s do an example chemistry problem, using the same principles.
We’ll look at each reactant, assume that it is used up completely, and calculate the
amount of one of the products.
Whichever reactant gives you the smallest amount of product is the limiting reactant.
Example 1: Here’s the combustion reaction of sucrose:
C12H22O11 + 12 O2 → 12CO2 + 11 H2O Start with 50.0 grams of sucrose
and 50.0 grams of oxygen.
Which is the limiting reagent?
Here’s our Strategy: First, we start with a Balanced Chemical Equation
Next, we calculate moles of product from each reactant
And finally, we see which is the smallest.
STEP 1: Start by confirming you have a balanced chemical equation.
If it isn’t balanced, you need to balance it before you can begin.
...There are the same number of Carbons on each side, same number of hydrogens, and the same
number of oxygens.
okay, this equation is balanced.
STEP 2: Find the molar mass of the reactants.
We’ll use this to convert everything from grams to moles.
That will let us us the balanced chemical equation to make conversion factors.
Molar mass of sucrose is 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol
50.0 g C12H22O11 (1 mol/342.34g) = 0.146 mol sucrose
Now we'll do oxygen.
Molar mass of oxygen is 2(16.00) = 32.00 g/mol. 50.0 g O2 (1 mol O2/32 g O2) = 1.56 mol O2.
STEP 3: Find how much product you would get if all of a reactant is used up.
Now that we have the reactants in moles, we can use the balanced chemical equation to
find the moles of CO2 they would produce if they were completely used up.
0.146 mol sucrose (12 mol CO2/1 mol sucrose) = 1.75 mol CO2
In other words, if all the sucrose reacted it would produce 1.75 mol CO2.
Now we do the same thing for oxygen.
1.56 mol O2 (12 mol CO2/12 mol O2) = 1.56 mol CO2
If all the O2 reacted, it would produce 1.56 mol CO2
The 50g of oxygen produced less CO2 than the 50g of sucrose,
so oxygen is the limiting reagent.
Next, let's find how many grams of CO2 will be made?
Since Oxygen is the limiting reagent, that will determine how much CO2 will be made.
Once all the O2 is used up, the reaction stops - even though there will be sucrose left over.
We found in part a that 1.56 mol of CO2 will be produced.
We will use the molar mass of CO2 to convert this to grams.
Molar mass of CO2 = 12.01 + 2(16.00) = 44.01 g/mol
1.56 mol CO2 (44.01 g CO2/ 1mol CO2) = 68.66 g CO2
Next, we'll find how much is leftover of the other reagent?
We made 1.56 mol of CO2.
How many moles of sucrose does that correspond to?
We’ll use the balanced chemical equation to figure this out.
1.56 mol CO2 (1 mol sucrose/12 mol CO2) = 0.13 mol sucrose
0.13 mol sucrose (342.34 g sucrose/1 mol sucrose) = 44.50 g sucrose
Careful - that’s not the answer.
That’s how much we used, and the question was how much is left.
To get that, subtract how much we used from the starting amount:
50.0 g sucrose - 44.50 g sucrose = 5.5 g of sucrose left over.
And finally, we'll find how much of the limiting reagent would you have to add to use up all the excess reagent?
In the last question, we found there were 5.5 g of sucrose left.
How much oxygen does that correspond to?
We’ll convert to moles of sucrose and then use the balanced chemical equation to find
out how much oxygen that means.
5.5 g sucrose (1 mol sucrose/ 342.34 g sucrose) = 0.016 mol sucrose
From the balanced chemical equation, we see that for every mole of sucrose,
we need 12 moles of oxygen.
So we have 0.016 mol sucrose (12 mol O2/ 1 mol sucrose)
= 0.193 mol O2 Use the molar mass of O2 to find out how many
grams that is.
0.193 mol O2 (32 g O2/ 1 mol O2) = 6.17 g O2
So it would take 6.17 g O2 to react with the excess sucrose.
Example 2: Here’s an example of WHY the limiting reagent idea is so useful.
Silver Nitrate reacts with Sodium Chloride to produce Silver Chloride
for photographic film.
The Silver nitrate is way more expensive than the sodium chloride, so this reaction is usually
performed with an excess of NaCl.
How many grams of AgNO3 are needed to produce 200 grams of AgCl?
First, we check that the chemical equation is balanced - and this equation is already balanced.
Next, we'll find the molar masses of AgCl and AgNO3 AgCl = 107.87 + 35.45 = 143.32 g/mol
AgNO3 = 107.87 + 14.01 + 3(16.00) = 169.88 g/mol
We started with 200.0 g AgCl (1 mol/143.32 g) = 1.395 mol AgCl
Using the Balanced chemical equation, 1.395 mol AgCl (1 mol AgNO3/1 mol AgCl)= 1.395 mol
AgNO3
1.395 mol AgNO3 (169.88 g/1 mol AgNO3) = 237.06 g AgNO3
In real life, we would measure out exactly 237.06g of AgNO3, and put NaCl in excess.
We’ll run out of AgCl, and have NaCl left over.
This way, we don’t waste any of the more expensive reagent.
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