Stoichiometry: the mass relationships between reactants and products. We will use the molar masses ( amount of

grams in one mole of a element or compound) from the periodic table

And the coefficients from the balanced equations to solve problems

When the following reaction occurs:Ba + Cl2 BaCl2

One atom of barium reacts with 1 molecule of chlorine to make one molecule of barium chloride.

It does not mean that 1 gram of barium reacts with 1 gram of chlorine to make 1 gram of barium chloride!

Remember- barium and chlorine are different sizes and have different masses!

We use stoichiometry to relate the atoms to masses- something we can easily measure.

Not only signify the ratio of atoms that react- also the ratio of moles of that substance involved.

Mole Ratio: a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction.

For example:

2Al2O3 4Al + 3O2

The mole ratio of Al2O3 to Al is 2:4, the mole ratio of Al2O3 to O2 is 2:3

You can use the mole ratio to convert from moles of one substance to moles of another.

Use the following reaction to answer the question: 4NH3 + 6NO 5N2 + 6H2O

1.If I have 20 moles of NH3, how many moles of N2 can I make?

2o mols NH3 5 mols N2 = 25 mols N2

4 mols NH3

Moles given

Coefficient wanted

Coefficient given

Moles wanted

If I want to make 2 moles of NaCl, How many mols of NaI do I need?2 mols

If I have 10 moles of Cl2, how many moles of I2 can I make?10 mols

If I want 15 mols of I2, how many mols of NaI do I need?30 mols

You can use molar mass of the substance to convert from moles to grams.

Remember, the molar mass on the periodic table is the number of grams in 1 mole.

How many grams in 1 mole of Na? 23 g

How many grams in 1 mole of NaCl?58.5 g

1 mol given

Molar mass of given

Grams of given

Coefficient of given

Coefficient of wanted Moles of

wanted

If I have 100 g of NaI, how many moles of I2 can I make? (molar mass of NaI is 150 g/mol)

100 g NaI 150 g

NaI

1 mol NaI

2 mol NaI

1 mol I20.033 mol I2

If I want to make 150 g of NaCl, how many moles of NaI do I need?

2.6 mols

Moles of given

1 mol wanted

Molar mass of wanted

Coefficient of wanted grams of

wantedCoefficient of given

If I have 10 moles of NaI, how many grams of I2 can I make?

10 moles NaI

2 moles NaI

1 mol I2

1 mol I2

254 g I2 = 1270 g

I2

If I need 15 moles of NaCl, how many grams of Cl2 do I need?

532 grams

grams of given

Molar mass of given

1 Mol given

Coefficient of given

Coefficient of wanted

Molar mass of wanted1 mole of wanted

= Grams of wanted

If I have 500 g of NaI, how many grams of I2 can I make?

2 mol NaI

1 mol I2

254 g I2= 423.3 g I2

1 mol I21 mol NaI150g

NaI

500 g NaI

If I want 150 g of NaCl, how many grams of I2 will I make?

325.6 g I2

Remember 1 mol = 22.4 L of a gas.

Are you LOST???

1. When you do a chemical experiment one of the reactants is going to be used up first.

2. The reactant that is used up first in a chemical experiment is called the limiting reactant

3. You can’t tell limiting reactants until you look at mole ratios

What is the limiting reactant in the picture?

4. How to find the limiting reactantA. Step 1: Balance equationB. Step 2: Convert both reactants to molesC. Step 3: Start with one reactant and find how

many grams of a product you will get (choose any product)

D. Step 4: Do the same with the other reactantE. Step 5: whichever gives less product is the

limiting reactant

1. Use the following equation to answer the next question

____ MnO2 + ____HCl ____MnCl2 + ____H2O + ____ Cl2

You have 5.47 grams of MnO2 and 9.861 grams of HCl. What is the limiting reactant?

A. Step 1: balance the chemical equation

____ MnO2 + ____HCl ____MnCl2 + ____H2O + ____ Cl21 4 1 2 2

B. Step 2: using the reactants, find the grams of a product stoichiometry. Essentially, you will be doing two stoichiometry problems. In this example I chose MnCl2 as the product to find grams of

5.47 g of MnO2

1 mol of MnO2

87 g of MnO2 1 mol of MnO2

1 mol of MnCl2 1 mol of MnCl2

125.84 g of MnCl2

7.91 g of MnCl2

9.861 g of HCl 1 mol of HCl

36.45 g of HCl

4 mol of HCl

1 mol of MnCl2 1 mol of MnCl2

125.84 g of MnCl2

8.511 g of MnCl2

C. Step 3: Compare the two numbers you found for your products. The lower of the two numbers is the limiting reactant. In other words, it is the reactant that will run out first.

MnO2 is the limiting reactant

1. Once the limiting reactant is found, we can use that value as the theoretical yield

What this tells us is that we should get this amount of product after we do the experiment.

2. The amount of product that actually forms during an experiment is called actual yield

It’s impossible to get 100% yields from our in class experiments.

1. To find out how close to the maximum yield you have you calculate a percent yield

2. Percent error tells you how far off you are from producing the maximum yield.

3. Percent yield and percent error should add up to 100%

Percent yield equation

Percent error equation

100 Yield lTheoretica

Yield Actualx

100 Yield lTheoretica

Yield Actual- lTheoreticax

Use the equation for the following problems

____Al + ____O2 ____Al2O3

1. What is the theoretical yield of Al2O3 if you start with 5.782 grams of Al?

A. Step 1: balance the chemical equation

____Al + ____O2 ____Al2O34 23

B. Convert 5.782 grams of Al to grams of Al2O3

using stoichiometry. Doing this will give you the theoretical yield. In actuality, you have been finding theoretical yields the whole unit.

5.782 g of Al 1 mol of Al

26.98 g of Al 4 mol of Al

2 mol of Al2O3 1 mol of Al

101.96 g of Al2O3

10.93 g of Al2O3

2. If the actual yield is 8.64 grams of Al2O3,what is the percent yield?

100 g 10.82

64.8x

g

= 79.9%

3. What is the percent error?

100 g 10.93

64.893.10x

gg

= 21%