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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