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1. Chemical Reactions:
An Introduction

2. 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

3. 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

5. 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

6. 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

7. Combustion of Methane
• methane gas burns to produce carbon
dioxide gas and liquid water
– whenever something burns it combines with O2(g)
CH4(g) + O2(g) → CO2(g) + H2O(l)
H
H
C
H
H
OO+
O
O
C + O
H H
1 C + 4 H + 2 O 1 C + 2 O + 2 H + O
1 C + 2 H + 3 O

8. 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
H
C
H
H
OO
+
O
O
C +
O
H H
OO
+
O
H H
+
1 C + 4 H + 4 O 1 C + 4 H + 4 O

9. 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

10. Symbols Used in Equations
• symbols used after chemical formula to
indicate state
– (g) = gas; (l) = liquid; (s) = solid
– (aq) = aqueous, dissolved in water

11. 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)

12. Balancing by Inspection
Count atoms of each element
a polyatomic ions may be counted as one
“element” if it does not change in the reaction
Al + FeSO4 → Al2(SO4)3 + Fe
1 SO4 3
b if an element appears in more than one
compound on the same side, count each
separately and add
CO + O2 → CO2
1 + 2 O 2

13. Balancing by Inspection
• 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

14. Examples
• 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
– 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)

15. Examples
• when magnesium metal burns in air it produces
a white, powdery compound magnesium oxide
– burning in air means reacting with O2
count the number of atoms of on each side
– count polyatomic groups as one “element” if on both
sides
– split count of element if in more than one compound
on one side
Mg(s) + O2(g) → MgO(s)
1 ← Mg → 1
2 ← O → 1

16. Examples
• when magnesium metal burns in air it produces
a white, powdery compound magnesium oxide
– burning in air means reacting with O2
pick an element to balance
– avoid element in multiple compounds
° find least common multiple of both sides & multiply each
side by factor so it equals LCM
Mg(s) + O2(g) → MgO(s)
1 ← Mg → 1
1 x 2 ← O → 1 x 2

17. Examples
• when magnesium metal burns in air it
produces a white, powdery compound
magnesium oxide
– burning in air means reacting with O2
± use factors as coefficients in front of compound
containing the element
if coefficient already there, multiply them together
Mg(s) + O2(g) → 2 MgO(s)
1 ← Mg → 1
1 x 2 ← O → 1 x 2

18. Examples
• when magnesium metal burns in air it produces
a white, powdery compound magnesium oxide
– burning in air means reacting with O2
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
2 ← O → 2

19. Examples
• 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
– identify the state of each chemical
ammonia(g) + oxygen(g) → nitrogen monoxide(g) + water(g)
write the equation in formulas
– identify diatomic elements
– identify polyatomic ions
– determine formulas
NH3(g) + O2(g) → NO(g) + H2O(g)

20. Examples
• Under appropriate conditions at 1000°C ammonia gas
reacts with oxygen gas to produce gaseous nitrogen
monoxide and gaseous water
count the number of atoms of on each side
– count polyatomic groups as one “element” if on both
sides
– split count of element if in more than one compound
on one side
NH3(g) + O2(g) → NO(g) + H2O(g)
1 ← N → 1
3 ← H → 2
2 ← O → 1 + 1

21. Examples
• Under appropriate conditions at 1000°C
ammonia gas reacts with oxygen gas to
produce gaseous nitrogen monoxide and
gaseous water
pick an element to balance
– avoid element in multiple compounds
° find least common multiple of both sides &
multiply each side by factor so it equals LCM
NH3(g) + O2(g) → NO(g) + H2O(g)
1 ← N → 1
2 x 3 ← H → 2 x 3
2 ← O → 1 + 1

22. Examples
• Under appropriate conditions at 1000°C
ammonia gas reacts with oxygen gas to
produce gaseous nitrogen monoxide and
gaseous water
± 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

23. Examples
• Under appropriate conditions at 1000°C ammonia gas
reacts with oxygen gas to produce gaseous nitrogen
monoxide and gaseous water
Recount
2 NH3(g) + O2(g) → NO(g) + 3 H2O(g)
2 ← N → 1
6 ← H → 6
2 ← O → 1 + 3
Repeat
2 NH3(g) + O2(g) → 2 NO(g) + 3 H2O(g)
2 ← N → 1 x 2
6 ← H → 6
2 ← O → 1 + 3

24. Examples
• Under appropriate conditions at 1000°C
ammonia gas reacts with oxygen gas to produce
gaseous nitrogen monoxide and gaseous water
´ Recount
2 NH3(g) + O2(g) → 2 NO(g) + 3 H2O(g)
2 ← N → 2
6 ← H → 6
2 ← O → 2 + 3

25. Examples
• Under appropriate conditions at 1000°C ammonia gas
reacts with oxygen gas to produce gaseous nitrogen
monoxide and gaseous water
Repeat
– 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) + 2.5 O2(g) → 2 NO(g) + 3 H2O(g)
2 ← N → 2
6 ← H → 6
2.5 x 2 ← O → 2 + 3