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Chapter 7: Chemical Reactions Section 7.1 Describing Reactions
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Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Jan 03, 2016

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Barry Wilkinson
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Page 1: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Chapter 7: Chemical Reactions Chapter 7: Chemical Reactions

Section 7.1

Describing Reactions

Section 7.1

Describing Reactions

Page 2: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Chemical ReactionsChemical Reactions

http://www.youtube.com/watch?v=Mx5JJWI2aaw&feature=related

http://www.youtube.com/watch?v=Mx5JJWI2aaw&feature=related

Page 3: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

EquationsEquations•

Examples: With words

carbon + oxygen -------> carbon dioxide

With formulas (a chemical equation) C + O2 ----------> CO2

Examples: With words

carbon + oxygen -------> carbon dioxide

With formulas (a chemical equation) C + O2 ----------> CO2

Page 4: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

The Law of Conservation of Mass

The Law of Conservation of Mass

conservation of mass clip conservation of mass clip

Page 5: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Balancing Equations Balancing Equations Look at this equation:

N2H4 + O2 ---> N2 + H2O•Are the numbers of nitrogen atoms, hydrogen, and oxygen

atoms the same on both sides?

NO! We need to balance the equation.

do this by changing the coefficients (the numbers that appear before the formulas).

When there is no other, a coefficient of 1 is assumed.

Look at this equation:N2H4 + O2 ---> N2 + H2O•Are the numbers of nitrogen atoms, hydrogen, and oxygen

atoms the same on both sides?

NO! We need to balance the equation.

do this by changing the coefficients (the numbers that appear before the formulas).

When there is no other, a coefficient of 1 is assumed.

Page 6: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

N2H4 + O2 N2 + H2ON2H4 + O2 N2 + H2O

N=

H =

O =

N=

H =

O =

N =

H =

O =

N =

H =

O =

When you add a coefficient of 2, you get this balanced equation:

N2H4 + O2 N2 + 2H2O

Page 7: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

More practice with balancing equations

More practice with balancing equations

Balance these three equations: Cu + O2 --------> CuO

H2O2 -------> H2O + O2

Mg + HCl ------> H2 + MgCl2

Balance these three equations: Cu + O2 --------> CuO

H2O2 -------> H2O + O2

Mg + HCl ------> H2 + MgCl2

Page 8: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

MolesMoles

••

Page 9: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

A mole is a lot of things!A mole is a lot of things!

To comprehend the enormous size of Avogadro’s number, 6.02 x 1023, here are some analogies.

Would a mole of rice grains fill our chemistry lab?

Would a mole of rice grains fit in our school?

1 mole of rice grams would cover all the land area of the whole world to a depth of 75 meters

To comprehend the enormous size of Avogadro’s number, 6.02 x 1023, here are some analogies.

Would a mole of rice grains fill our chemistry lab?

Would a mole of rice grains fit in our school?

1 mole of rice grams would cover all the land area of the whole world to a depth of 75 meters

Page 10: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

1 mole of rice has more grains than the number grains of all rice grown since the beginning of time!

1 mole of rice has more grains than the number grains of all rice grown since the beginning of time!

Page 11: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

1 mole of watermelon seeds:1 mole of watermelon seeds:

Would be found in a watermelon slightly larger than the moon!

Would be found in a watermelon slightly larger than the moon!

Page 12: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

1 mole pennies divided equally between every person on earth:

1 mole pennies divided equally between every person on earth:

Each person would receive 1 x 1014 pennies.

Personal spending at the rate of $1 million per day would use up each person’s wealth in just under 3000 years!

Life would not be comfortable. The surface of our planet would be buried in copper coins to a depth of about 420 m.

Each person would receive 1 x 1014 pennies.

Personal spending at the rate of $1 million per day would use up each person’s wealth in just under 3000 years!

Life would not be comfortable. The surface of our planet would be buried in copper coins to a depth of about 420 m.

Page 13: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Each human being has about 60 million body cells (6 x 1013)

Each human being has about 60 million body cells (6 x 1013)

Assume that the earth’s population is 6 billion

(6 x 109), the total number of living body cells on the earth at the present time is 3.6 x 1023 or a little more than half a mole!

Assume that the earth’s population is 6 billion

(6 x 109), the total number of living body cells on the earth at the present time is 3.6 x 1023 or a little more than half a mole!

Page 14: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Molar MassMolar Mass

CO2: carbon (12 g) + oxygen (2g x 16g) = 44g CO2: carbon (12 g) + oxygen (2g x 16g) = 44g

Page 15: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Mole-Mass ConversionsMole-Mass Conversions

Example: You have 55 grams of CO2. How many moles of

CO2 do you have?

Example: You have 55 grams of CO2. How many moles of

CO2 do you have?

Page 16: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

More calculations….More calculations….

Page 17: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Examples to work Examples to work

How much oxygen is needed to make 144 grams of water?

two things are needed to work this problem ~ a balanced equation and molar masses.

first step is to find out how many moles of water we are trying to make.

144 g of H2O x 1 mol H2O = 8 mol H2O

18 g H2O

How much oxygen is needed to make 144 grams of water?

two things are needed to work this problem ~ a balanced equation and molar masses.

first step is to find out how many moles of water we are trying to make.

144 g of H2O x 1 mol H2O = 8 mol H2O

18 g H2O

Page 18: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Next step….Next step…. Next step - go to the balanced equation for a

ratio of moles of oxygen to moles of water.

2H2 + O2 --------> 2H2O

1 mol O2 OR 2 mol H2O

2 mol H2O 1 mol O2

Next step - go to the balanced equation for a ratio of moles of oxygen to moles of water.

2H2 + O2 --------> 2H2O

1 mol O2 OR 2 mol H2O

2 mol H2O 1 mol O2

Page 19: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Next step….Next step…. convert moles of water you are trying to

make and then to moles of oxygen needed to make that amount of water.

8 mol H2O x 1 mol O2 = 4 mol O2

2 mol H2O

convert moles of water you are trying to make and then to moles of oxygen needed to make that amount of water.

8 mol H2O x 1 mol O2 = 4 mol O2

2 mol H2O

Page 20: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

Finally…..Finally…..

Convert the moles of oxygen needed to the mass of oxygen needed.

4 mol O2 x 32 grams = 128 grams

1 mol O2

So, 128 grams of oxygen must be used in order to produce 144 grams of water.

Convert the moles of oxygen needed to the mass of oxygen needed.

4 mol O2 x 32 grams = 128 grams

1 mol O2

So, 128 grams of oxygen must be used in order to produce 144 grams of water.

Page 21: Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions.

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