## How do I find the limiting reagent given a chemical equation?

Example:

2CH2Cl2+3O2=>2CO2+2H2O+2Cl2

For the balanced equation shown above, what would be the limiting reagent if 76.0 grams of CH2Cl2 were reacted with 69.8 grams of O2?

There are a few steps that are necessary to find the limiting reagent.

1. The first step is calculating the molar mass of each chemical compound.

We do this by looking at the atomic number of each element in the compound:

C = 12

H = 1

Cl = 35.5

O = 16

Then we multiply there atomic number by how many atoms there are of this element in the compound:

12 x 1 = 12

1 x 2 = 2

35.5 x 2 = 70

16 x 2 = 32

Next we add this calculations together:

12+2+70 = 84 grams/mole

16 = 16 grams/mole

2. So, now that we know the molar mass of our compounds we need to convert the amount of grams given in the question into moles. Lets look at the question again.

For the balanced equation shown above, what would be the limiting reagent if 76.0 grams of CH2Cl2 were reacted with 69.8 grams of O2?

There are 76.0 grams of CH2Cl2 and 69.8 grams of O2. So, in order to convert the grams into moles we have the divide the amount of grams by the molar mass.

76/84 = .904 moles

69.8/16 = 4.362 moles

3. Finally, we look at the equation:

2CH2Cl2+3O2=>2CO2+2H2O+2Cl2

Lets only look at the first part:

2CH2Cl2+3O2

The coefficients are 2 and 3, they are there to show the ratio. So, this means that for every 2 moles of CH2Cl2 there are 3 moles of O2. 2/3

Now that we know the ratio we can set up a proportion.

.904 2

4.362 x 3

We cross multiply and end up with:

.904 x 3 = 3.612

4.362 x 2 = 8.724

The less product is the one that is the limiting reagent. So because 3.612 is less then 8.724 we know that CH2Cl2 is the limiting reagent because we got 3.612 by multiplying .904 by 3 and .904 is the number of moles of CH2Cl2 that we had.