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Introductory Chemistry: A Foundation THIRD EDITION by Steven S. Zumdahl University of Illinois
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2. Unit 2 Section 2 Chemical Reactions
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3. Do-Now
In our demonstration - 1. What were the two reactants that were combined? 2. What was the new product(s) produced that we observed as a class? 3. Was this an example of a physical change or a chemical change? Why?

4. Chemical Reactions Examples of Chemical Reactions Rusting of iron
Burning (combustion) of wood
Cooking
Photosynthesis
6CO2 + 6H2O + Energy --> C6H12O6 + 6O2

5. Chemical reactions are like sentences.
Letters 
Chemical Symbols
Words 
Chemical Formulas
Sentences 
Chemical Equations
Symbols  elements
Formula  compound
Equations  chemical reactions

6. Chemical reaction
a reaction that occurs when the atoms of one of more substances are rearranged to form different substances
a chemical change
Evidence of a chemical reaction
- Temperature change
- Color change
- Odor produced
- Formation of bubbles
- Appearance of a solid

7. Chemical Equations
2H2(g) O2(g) H2O (l)
Reactants: elements or compounds to the left of the arrow that combine together in a chemical reaction
Products: elements or compounds to the right of the arrow that are produced in a chemical reaction
Coefficient: whole number before a chemical formula
+ : represents “and”
: direction reaction progresses, yields or produces

8. C(s) O2(g) CO2 (g)
physical states of reactants and products are indicated by:
(s) = solid:    C(s)
(g) = gas:    CO2(g) 
(l) = liquid:    H2O(l)
(aq) = aqueous, dissolved in water:
NaCl(aq) is a salt water solution
(cr) = crystalline
= precipitate forms

9. Word and Formula Equations
A word equation is an equation represented by words.
Example: solid sodium plus chlorine gas reacts to produce sodium chloride
Skeleton equation represents the reactants and products of a chemical reaction by their formulas.
Example: Na(s) + Cl2(g) NaCl(s)

10. Balancing Chemical Equations
Law of conservation of mass - in a chemical reaction matter is neither created nor destroyed
Chemical equation - the number of atoms of each reactant must equal the number of atoms of each product
Equations must be balanced using coefficients
balanced chemical equation – shows that each side of the equation has the same number of atoms of each element and mass is conserved
Na(s) Cl2(g) NaCl(s) 2 2

11. Steps for Balancing Equations
Assemble the correct formulas for all the reactants and products, using “+” and “→”
2) Identify diatomic elements
3) Count the number of atoms of each type appearing on the reactant side and the product side
4) Balance the elements one at a time by adding coefficients in front of a formula to make number of atoms equal on both sides of equation
5) Check to make sure equation is balanced and coefficients are in lowest possible ratio

12. Practice Balancing Equations
1) Fe + S Fe2S3 2) KClO3 KCl + O2 3) Ca(OH)2 + HCl CaCl2 + H2O

13. Sample problems – Balancing Equations
Word Equation Water and carbon yield hydrogen gas and carbon monoxide Skeleton Equation Balanced Equation

14. Sample problems – Balancing Equations
Word Equation Magnesium chloride and potassium react to form potassium chloride and magnesium Skeleton Equation Balanced Equation

15. CH4(g) + O2(g)  CO2(g) + H2O(l)
Combustion of Methane
Word Equation
methane gas burns to produce carbon dioxide gas and liquid water
whenever something burns it combines with O2(g)
Skeleton Equation
CH4(g) + O2(g)  CO2(g) + H2O(l) H C O + 6

16. Combustion of Methane Balanced
to show the reaction obeys the Law of Conservation of Mass it must be balanced
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l) H C O +
1 C H O
1 C H O
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17. Water and iron react to form iron III oxide and hydrogen
Propane (C3H8) burns to form carbon dioxide and water

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Chemical Reactions
Reactions involve chemical changes in matter resulting in new substances
Reactions involve rearrangement and exchange of atoms to produce new molecules
Elements are not transmuted during a reaction
Reactants  Products
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21. Evidence of Chemical Reactions
a chemical change occurs when new substances are made
visual clues (permanent)
color change, precipitate formation, gas bubbles, flames, heat release, cooling, light
other clues
new odor, permanent new state
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Evidence of Chemical Reactions:
Color Change
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Evidence of Chemical Reactions
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Chemical Equations
Shorthand way of describing a reaction
Provides information about the reaction
Formulas of reactants and products
States of reactants and products
Relative numbers of reactant and product molecules that are required
Can be used to determine weights of reactants used and of products that can be made
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Conservation of Mass
Matter cannot be created or destroyed
In a chemical reaction, all the atoms present at the beginning are still present at the end
Therefore the total mass cannot change
Therefore the total mass of the reactants will be the same as the total mass of the products
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Writing Equations
Use proper formulas for each reactant and product
proper equation should be balanced
obey Law of Conservation of Mass
all elements on reactants side also on product side
equal numbers of atoms of each element on reactant side as on product side
balanced equation shows the relationship between the relative numbers of molecules of reactants and products
can be used to determine mass relationships
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27. Symbols Used in Equations
symbols used after chemical formula to indicate physical state
(g) = gas; (l) = liquid; (s) = solid
(aq) = aqueous, dissolved in water
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28. Sample – Recognizing Reactants and Products
when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
burning in air means reacting with O2
Metals are solids, except for Hg which is liquid
write the equation in words
identify the state of each chemical
magnesium(s) + oxygen(g) magnesium oxide(s)
write the equation in formulas
identify diatomic elements
identify polyatomic ions
determine formulas
Mg(s) + O2(g)  MgO(s)
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29. Balancing Chemical Equations
Count atoms of each element
polyatomic ions may be counted as one “element” if it does not change in the reaction
Al + FeSO4 Al2(SO4)3 + Fe
1 SO
if an element appears in more than one compound on the same side, count each separately and add
CO + O2  CO2
O 2
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30. Balancing Chemical Equations
Pick an element to balance
avoid elements from 1b
Find Least Common Multiple and factors needed to make both sides equal
Use factors as coefficients in equation
if already a coefficient then multiply by new factor
Recount and Repeat until balanced
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31. Example 1: Magnesium metal burns in air
when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
burning in air means reacting with O2
write the equation in words
magnesium(s) + oxygen(g) magnesium oxide(s)
write the equation in formulas - determine formulas
Mg(s) + O2(g)  MgO(s)
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32. Example 1: Magnesium metal burns in air
count the number of atoms of on each side
Mg(s) + O2(g)  MgO(s)
1  Mg 1
2  O  1
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33. Example 1: Magnesium metal burns in air
pick an element to balance
- avoid element in multiple compounds
Mg(s) + O2(g)  MgO(s)
1  Mg 1
1 x 2  O  1 x 2
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34. Example 1: Magnesium metal burns in air
Mg(s) + O2(g)  2 MgO(s)
1  Mg 1
1 x 2  O  1 x 2
Use factors as coefficients in front of compound containing the element
if coefficient already there, multiply them together
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35. Example 1: Magnesium metal burns in air
Recount
Mg(s) + O2(g)  2 MgO(s)
1  Mg 2
2  O  2
Repeat
2 Mg(s) + O2(g)  2 MgO(s)
2 x 1  Mg 2
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36. Example 2 NH3(g) + O2(g)  NO(g) + H2O(g)
Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
write the equation in words
ammonia(g) + oxygen(g) nitrogen monoxide(g) + water(g)
write the equation in formulas
NH3(g) + O2(g)  NO(g) + H2O(g)
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37. Examples NH3(g) + O2(g)  NO(g) + H2O(g) 1  N 1 3  H  2
count the number of atoms of on each side
NH3(g) + O2(g)  NO(g) + H2O(g)
1  N 1
3  H  2
2  O  1 + 1
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38. Examples NH3(g) + O2(g)  NO(g) + H2O(g) 1  N 1 2 x 3  H  2 x 3
pick an element to balance - avoid element in multiple compounds
NH3(g) + O2(g)  NO(g) + H2O(g)
1  N 1
2 x 3  H  2 x 3
2  O  1 + 1
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39. Examples 2 NH3(g) + O2(g)  NO(g) + 3 H2O(g) 1  N 1
Use factors as coefficients in front of compound containing the element
2 NH3(g) + O2(g)  NO(g) + 3 H2O(g)
1  N 1
2 x 3  H  2 x 3
2  O  1 + 1
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40. Examples Recount 2 NH3(g) + O2(g)  NO(g) + 3 H2O(g) 2  N 1
Repeat
2 NH3(g) + O2(g) 2 NO(g) + 3 H2O(g)
2  N 1 x 2
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41. Examples Recount 2 NH3(g) + O2(g)  2 NO(g) + 3 H2O(g) 2  N 2
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42. Examples Repeat 2 NH3(g) + 2.5 O2(g) 2 NO(g) + 3 H2O(g) 2  N 2
A trick of the trade, when you are forced to attack an element that is in 3 or more compounds – find where it is uncombined. You can find a factor to make it any amount you want, even if that factor is a fraction!
We want to make the O on the left equal 5, therefore we will multiply it by 2.5
2 NH3(g) O2(g) 2 NO(g) + 3 H2O(g)
2  N 2
6  H  6
2.5 x 2  O  2 + 3
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43. Examples
Multiply all the coefficients by a number to eliminate fractions:
2 x [2 NH3(g) O2(g) 2 NO(g) + 3 H2O(g)]
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
4  N 4
12  H  12
10  O  10
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