Section 12.2

Chemical Calculations

1

Chapter 12

Stoichiometry

Mole Ratios

2

These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine are needed to react with 5 moles of sodium (without any sodium left over)?

Na + Cl2 NaCl2 mol 1 mol5 mol x

2Cl mol5.22

15

x

22

Step 1: Write the chemical equation.Step 2: Balance the equation.

Step 3:Step 4:

Step 5:

(Write the mole ratio)(Write the given)

Mole - Mole Conversions

How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?

3

2 Na + Cl2 2 NaCl1 mol 2 mol

2.6 mol x

NaCl mol2.51

26.2

x

Example:

Mole - Mole Conversions

How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?

4

Example:

Al2O3 ® Al + O2

2 mol 3 mol

2O mol01.52

334.3

x

3.34 mol x

3 4 2

Mole - Mass Conversions

Most of the time in chemistry, the amounts

are given in grams instead of moles

We still go through moles and use the mole ratio, but now we also use molar mass to get to grams

5

Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride?

6

Na + Cl2 NaCl222 mol 1 mol5 mol x

2Cl mol5.22

15

x

Change moles to grams:No. of moles X molar mass2.5 X 71 = 178 g Cl2

Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

7

Example:

I2 + Al ® AlI3 233 mol 2 mol

x 0.5 mol

2

2I mol75.02

5.03

x

Change moles to grams:No. of moles X molar mass0.75 X 254 = 191 g I2

Mass – Mole Conversions

We can also start with mass and convert to

moles of product or another reactant

We use molar mass and the mole ratio to get

to moles of the compound of interest

8

Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water

9

Example:

2 C2H6 + 7 O2 4 CO2 + 6 H2O 2 mol 6 mol

x 10 gBut before making cross multiplication, change 10 grams to moles

10 / 18.02 = 0.555 mol

62HC 185.06

2555.0molx

Mass – Mass ConversionExample: Calculate how many grams of ammonia are

produced when you react 2.00g of nitrogen with excess hydrogen.

10

N2 + H2 NH31 mol 2 mol

2 g x

3 2

But before making cross multiplication, change 2 grams to moles2 / 28 = 0.0714 mol

3NH mol143.01

20714.0

x

Change moles to grams:No. of moles X molar mass0.143 X 17.03 = 2.44 g NH3

Calculate how many grams of oxygen are required to make 10.0 g of aluminum oxide

11

Example:

Al2O33 4 2Al + O2®3 mol 2 mol

x 10 gBut before making cross multiplication, change 10 grams to moles

10 / 102 = 0.098 mol

2O mol147.02

3098.0

x

Change moles to grams:No. of moles X molar mass0.147 X 32 = 4.7 g O2

• If 3.84 moles of C2H2 are burned, how many

moles of O2 are needed?

12

(9.6 mol)

Example:

2 C2H2 + 5 O2 ® 4 CO2 + 2 H2O2 mol 5 mol

3.84 mol x

2O mol6.92

584.3

x

• How many moles of C2H2 are needed to produce

8.95 mole of H2O?(8.95 mol)

Example:

2 C2H2 + 5 O2 ® 4 CO2 + 2 H2O2 mol 2 mol

x 8.95 mol

22HC mol95.82

295.8

x

13

• If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?(4.94 mol)

Example:

2 C2H2 + 5 O2 ® 4 CO2 + 2 H2O

2 mol 4 mol

2.47 mol x

2CO mol94.42

447.2

x

14

Mass – Mass Problem:

15

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

4 Al + 3 O2 2Al2O3

4 mol 2 mol 6.5 g x

But before making cross multiplication, change 6.5 grams to moles

6.5 / 27 = 0.241 mol Al

32OAl mol121.04

2241.0

x

Change moles to grams:No. of moles X molar mass0.121 X 102 = 12.3 g Al2O3

Another example:

2 Fe + 3 CuSO4 ® Fe2(SO4)3 + 3 Cu

16

If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form?

2 mol 3 mol10.1 g x

before making cross multiplication, change 10.1 grams to moles 10.1 / 55.8 = 0.181 mol Fe

Cu mol272.02

3181.0

x

Change moles to grams:No. of moles X molar mass0.272 X 63.5 = 17.3 g Cu

Volume – Volume Calculations:

How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ?

17

CH4 + 2 O2 ® CO2 + 2 H2O1 mol 2 mol

x 17.5 L before making cross multiplication, change 17.5 L to moles 17.5 / 22.4 = 0.781 mol O2

4CH mol391.02

1781.0

x

Change moles to liters:No. of moles X 22.40.391 X 22.4 = 8.76 L CH4

Section 12.3

18

Limiting Reactant: Cookies1 cup butter, 1/2 cup white

sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate

chipsMakes 3 dozen

If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?

What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?

What if we only had one egg, could we make 3 dozen cookies?

Limiting Reactant

Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant.

That reactant is said to be in excess (there is too much). The other reactant limits how much

product we get. Once it runs out, the reaction s. This is called the limiting reactant.

Limiting Reactant To find the correct answer, we have to try all

of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one.

The lower amount of a product is the correct answer.

The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!

Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

Example 10.0g of aluminum reacts with 35.0 grams of chlorine

gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?

2 Al + 3 Cl2 2 AlCl3 Start with Al:

Now Cl2:

2 mol 2 mol 10 g x

But before making cross multiplication, change 10 grams to moles10 / 27 = 0.370 mol

3CA mol370.02

2370.0llx

3 mol 2 mol 35 g x

But before making cross multiplication, change 35 grams to moles35 / 71 = 0.493 mol

3CA mol329.03

2493.0llx

0.370 x 133.5 = 49.4 g

0.329 x 133.5 = 43.9 g

LR Example Continued

We get 49.4 g of aluminum chloride from the given amount of aluminum, but only 43.9 g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 33.6 g of chlorine is used up, the reaction comes to a complete .

24

If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of product (cuprous sulfide) will be formed?

Practice

25

15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

Finding the Amount of Excess

By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

Can we find the amount of excess potassium in the previous problem?

27

Practice15.0 g of potassium reacts with 15.0 g of iodine. If 19.6 g of potassium iodide is produced, find the limiting and excess reactants then how much of the excess amount is left over?

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.

Limiting Reactant: Recap1. You can recognize a limiting reactant

problem because there is MORE THAN ONE GIVEN AMOUNT.

2. Convert ALL of the reactants to the SAME product (pick any product you choose.)

3. The lowest answer is the correct answer.4. The reactant that gave you the lowest

answer is the LIMITING REACTANT.5. The other reactant(s) are in EXCESS.6. To find the amount of excess, subtract the

amount used from the given amount.7. If you have to find more than one product,

be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

Yield The amount of product made in a

chemical reaction. There are three types Actual yield- what you get in the lab

when the chemicals are mixed Theoretical yield- what the balanced

equation tells you should make. Percent yield = Actual x 100

% Theoretical

Percent Yield To determine percentage yield, you

need two pieces of information:

1)Theoretical yield =

2)Actual yield =

Comes from the stoichiometry

Comes from experimental data

Percent Yield = Actual Yield

Theoretical Yield

100%

Example 6.78 g of copper is produced when

3.92 g of Al are reacted with excess copper (II) sulfate.

2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu What is the actual yield? What is the theoretical yield? What is the percent yield? If you had started with 9.73 g of Al,

how much copper would you expect?

Details Percent yield tells us how “efficient” a

reaction is. Percent yield can not be bigger than

100 %.

35

12.75 moles of NaClO3 will produce how many grams of

O2? Calculate the percent yield of oxygen if 20 grams is

produced. NaClO3 NaCl + O2