1. Coach Hyde Physical Science Unit 5
Chemical Equations
Coach Hyde Physical Science Unit 5

2. The Law of Conservation of Matter: states that for any system closed to all transfers of matter and energy (both of which have mass), the mass of the system must remain constant over time, as system mass cannot change quantity if it is not added or removed
Soooo….. Matter cannot be created nor destroyed it can only change form
(this is the basic principle of all conservation laws such as energy, matter, mass etc…
Thank you Sir Antoine Lavoisier

3. (s): Solid stated of matter (l): Liquid State of matter
Something to discuss before we get started
(s): Solid stated of matter
(l): Liquid State of matter
(g): gaseous State of Matter
(aq): compound is in aqueous
solution with H2O

4. Balancing Chemical Equations

5. Reactants = The starting substances (things that react)
Products = The final substances (produced) after the reaction

6. Exothermic Reaction: occurs when the temperature of a system increases due to the evolution of heat. This heat is released into the surroundings,
A system of reactants that absorbs heat from the surroundings in an Endothermic Reaction. How do you think this can be noticed/determined?

7. If a chemical equation does not obey the law of
conservation of mass the equation is said to be what?
NOT BALANCED
So Let’s look at the steps we need to take to
BALANCE chemical equations
Let’s work with the following equation:
Fe O2  Fe2O3

8. Step 1. Create a RAP table
(what’s a RAP table ??)
A table that shows us what atoms are present in this
reaction, how many there are and are they reactants
or products?
For example:
Fe O2  Fe2O3
#R atom #P Fe O

9. Rule 2. Go to the first atom that’s not balanced and balance it!
Since Fe atoms are not balanced what do we need to do to balance it?
Right! Multiply it by 2 (Only multiply)
#R atom #P
1 Fe 2
2 O 3 2x

10. In step 2 we balanced the number of Fe atoms by multiplying the reactant side by 2. This now becomes the new coefficient in the chemical equation.
Modify the equation to reflect the change
#R atom #P
1 Fe 2
2 O 3
2Fe O2  Fe2O3 2x
Are all atoms balanced?

11. 3. Move to the next unbalanced atom. What is it?
#R atom #P Fe O
How can we balance the
Oxygen?
Multiply Reactants by 3 and Products by 2
Adjust the equation to reflect your
changes
#R atom #P Fe
3x O x2
2Fe O2  2Fe2O3
But notice that by changing Oxygen we also
Changed Iron. We need to go back and fix this.

12. 4. Write out the updated RAP table. How can we
Balance the Iron?
#R atom #P Fe O
Sure! Multiply the # of
Reactant Fe atoms by 2 ! 2x
Re-write the equation reflecting
The new changes you’ve made.
4Fe O2  2Fe2O3
Do we have a balanced Chemical Equation now?
Yes we do!

13. Polyatomics #R atom #P 2 Ag 2 2 NO3 2 Mg 1 2 Cl 2
When an equation has Polyatomics in it, such as in this
Balanced chemical equation
2AgNO3 + MgCl2  2AgCl + Mg(NO3)2
And the polyatomic appears on BOTH the reactant and product
Side of the equation Count the polyatomic as an “ATOM”
So the above reactant atoms would be:
#R atom #P Ag
NO3 2
Mg 1
2 Cl 2
If the same polyatomic does not
Appear on both sides break the
Polyatomic down into atoms!

14. Let's try solving some Problems

15. Is this equation balanced?
NaOH + CaBr2  Ca(OH)2 + NaBr
What atoms do we have in
This equation?
Count atoms & Start
the RAP table
#R atom #P NA OH Ca Br
2) Do the #Reactant atoms = the # of Product atoms?
3) So pick the 1st unbalanced atom & begin balancing

16. We’ll start with balancing Hydroxide
NaOH + CaBr2  Ca(OH)2 +NaBr
How can we make both Hydroxides equal?
#R atom #P Na
OH 2 Ca Br
Sure we’ll multiply #R OH by 2
Next step> rewrite the modified eqn. 2x
2NaOH + CaBr2  Ca(OH)2 + NaBr
Hydroxide is now balanced so let’s move to the next
Unbalanced atom, which is? …

17. What can we do to balance the Bromine?
Sure! Multiply the #P Bromine by 2
Now adjust the table to reflect
The changes and then rewrite the
Eqn.
#R atom #P
Na 1
OH 2 Ca
Br 1 x2
2NaOH + CaBr2  Ca(OH)2 + 2NaBr

18. Let’s update the RAP table with the new #’s
Based on our updated equation.
2NaOH + CaBr2  Ca(OH)2 + 2NaBr
#R atom #P
2 Na OH Ca Br
Are we now balanced?
Sure!

19. Ok Try Balancing this equation:
C2H6 + O2  CO2 + H2O
#R atom #P C H O
Step 1. Total up the atoms
Step 2. Balance the #P Carbon
#R atom #P C H O
& Re-write the equation
C2H6 + O2  2CO2 + H2O
Are we done?

20. Step 3. Carbons are balanced now but Hydrogen isn’t.
So, balance Hydrogen atoms next
#R atom #P C H O
Multiply #P Hydrogen by 3 x3
Step 4. Re-write the eqn. &
Retotal the number of atoms
C2H6 + O2  2CO H2O
#R atom #P C H O
Carbon and Hydrogen are now
balanced but oxygen isn’t.

21. Step 5. To balance Oxygen multiply O by 3½
#R atom #P C H O
Step 6. Re-write the eqn. &
Retotal the number
of atoms
3 ½ x
C2H ½ O2  2CO H2O
It looks like we’re balanced. But, are we?
No! We can’t have 3 ½ Oxygen molecules! Only whole
Numbers are allowed.
So what do we need to do to fix this?

22. Step 7. Let’s clean this up by Multiply everything by 2
C2H ½ O2  2CO H2O
x 2
2C2H O2  4CO H2O
#R atom #P C H O
Step 8. Retotal #R and the
#P atoms
Are we balanced?
YES!

23. Try this problem NH4OH + FeCl3  Fe(OH)3 + NH4Cl
Start here. Recognize we
Have polyatomics but they
Appear on both sides of the
Equation.
#R atom #P
NH4 1
OH 3 Fe Cl
OK … Now finish it up

24. Answer to previous problem
3NH4OH + FeCl3  Fe(OH)3 + 3NH4Cl

25. Types of Reactions You need to be able to identify each type.
Synthesis reactions
Decomposition reactions
Single displacement reactions
Double displacement reactions
Combustion reactions
You need to be able to identify each type.

26. 1. Synthesis
Example C + O2 O C +  O C
General: A + B  AB

27. Ex. Synthesis Reaction

28. Practice Predict the products. Na(s) + Cl2(g)  Mg(s) + F2(g) 
Al(s) + F2(g)  2 2
NaCl(s)
MgF2(s) 2 3 2
AlF3(s)
Now, balance them.

29. 2. Decomposition
Example: NaCl  Cl Na Cl + Na
General: AB  A + B

30. Ex. Decomposition Reaction

31. Went from neutral (0) to (+2) Went from (+2) to Neutral (0)
3. Single Displacement
Example: Zn + CuCl2
Zn was oxidized
Went from neutral (0) to (+2)  Zn Cl Cu + Cl Zn Cu +
Cu was reduced
Went from (+2) to Neutral (0)
General: AB + C  AC + B

32. Ex. Single Replacement Reaction

33. Single Replacement Reactions
Write and balance the following single replacement reaction equation:
Zn(s) HCl(aq)  ZnCl2 + H2(g) 2
NaCl(s) + F2(g)  NaF(s) + Cl2(g)
Al(s)+ Cu(NO3)2(aq) 2 2 2 3 3
Cu(s)+ Al(NO3)3(aq) 2

34. 4. Double displacement General: AB + CD  AD + CB Example: MgO + CaS 

35. Double Replacement Reactions
Think about it like “foil”ing in algebra, first and last ions go together + inside ions go together
Example:
AgNO3(aq) + NaCl(s)  AgCl(s) + NaNO3(aq)
Another example:
K2SO4(aq) + Ba(NO3)2(aq)  KNO3(aq) + BaSO4(s) 2

36. Practice Predict the products. HCl(aq) + AgNO3(aq) 
CaCl2(aq) + Na3PO4(aq) 
Pb(NO3)2(aq) + BaCl2(aq) 
FeCl3(aq) + NaOH(aq) 
H2SO4(aq) + NaOH(aq) 
KOH(aq) + CuSO4(aq) 

37. 5. Combustion Reactions
Combustion reactions - a hydrocarbon reacts with oxygen gas.
This is also called burning!!!
In order to burn something you need the 3 things in the “fire triangle”:
1) Fuel (hydrocarbon) 2) Oxygen 3) Something to ignite the reaction (spark)

38. Combustion Reactions In general: CxHy + O2  CO2 + H2O
Products are ALWAYS
carbon dioxide and water. (although incomplete burning does cause some by-products like carbon monoxide)
Combustion is used to heat homes and run automobiles (octane, as in gasoline, is C8H18)

39. Combustion
Example
C5H O2  CO2 + H2O
Write the products and balance the following combustion reaction:
C10H O2  8 5 6

40. Mixed Practice State the type & predict the products. BaCl2 + H2SO4 
C6H12 + O2 
Zn + CuSO4 
Cs + Br2 
FeCO3 