1. Why do chemical equations need to be balanced?
Today we are going to learn how to balance chemical equation!
Why do chemical equations need to be balanced?
Because…in a chemical equations, the mass of the products must equal the mass of the reactants
This is called the Law of Conservation of Mass

2. Chemical reaction: takes place when new products are made from reactants Possible signs of a chemical reaction: bubbling, temperature change, color change, smoke, or change in odor

3. Definitions: Reactants – Starting materials Products – Resulting substance(s) Chemical equation – Describes what happens in a chemical reaction, it identifies products, reactants, formulas of the participants, phases of the participants (solid, liquid, or gas) and amount of each substance

4. States of matter that can be included in the chemical equation: (g) = gas (s) = solid (l) = liquid (aq) = solution in water

5. Other important information: single arrow (→) = yields or produces, shows the direction of action delta ( Δ ) = shows that heat has been added double arrow ( ↔ ) = shows that the reaction is reversible or can go in both directions

6. Mini Math Lesson: Distributive Property
2(3y + 4) = 6y + 8
2 x 3 = x 4 = 8
You are distributing the “2”
The “2” in this case is called the “coefficient”

7. Mini Math Lesson: Distributive Property
3(4z + 6) =
12z + 18
10(5v + 3) =
50v + 30
In balancing chemical equations, it works the same way…

8. (reactants) (products)
Learning how to count atoms! Let’s use the equations for water: H2 + O  H2O 1
Equation is given to you:
H  H201 2 2
This part you have to write under the equation:
____ H ____ H
____ O ____ O
(reactants) (products)
Imaginary! 1 1

9. Learning how to count atoms with coefficients!1
Equation is given to you:
2 H  2 H201 4 4
This part you have to write under the equation:
____ H ____ H
(2 x 2) = (2 x 2) = 4
____ O ____ O
(2 x 1) = (2 x 1) = 2
(reactants) (products)
Imaginary! 2 2

10. It is still balanced, even though I’ve added coefficients!
Balanced chemical equations:
H2 + 0  H20 4
These are subscript numbers – they tell you how many of each atom there are (represented here with a green circle)
____ H ____ H
____ O ____ O
(reactants) (products) 2 2 1 1
2 H  2 H20
These are coefficients – numbers that come before the molecule and need to be distributed to the whole molecule (represented here with a yellow circle)
____ H ____ H
____ O ____ O
(reactants) (products) 4 4
It is still balanced, even though I’ve added coefficients! 2 2

11. Steps to balancing a chemical equation:
Step 1 – Count the number of atoms in the unbalanced chemical equation Step 2 – Add one coefficient at a time
Step 3 – Add a second coefficient where/when needed
Step 4 – Continue to add new coefficients or change the coefficients as needed until equation is balanced

12. Let’s try an example together! Step 1: Count the atoms
SnO2 + H2  Sn + H20
____ Sn ____ Sn
____ H ____ H
____ O ____ O
(reactants) (products) 1 1 2 2 2 1 2 1 4
Uh ohhhhhhhh! The oxygens aren’t balanced!

13. (reactants) (products)
SnO2 + H2  Sn + H20
____ Sn ____ Sn
____ H ____ H
____ O ____ O
(reactants) (products) 1 1 2 2 2 1
Since this is the original equation, we cannot decrease the number of oxygens so that both will be “1”
Instead, our only option is to increase the number of oxygens.
For example, both sides can be increased to “2” each, or “4” each, etc…

14. But it’s still not balanced, but we are 1 step closer!
What do we do?! We have to balance the equation! Step 2: ADD COEFFICIENTS!
We need “2 O” on the reactants side (left) and “2 O” on the products side (right)
__ SnO2 + __ H2  __ Sn + __ H20 2 2 1
____ Sn ____ Sn
____ H ____ H
____ O ____ O
(reactants) (products) 1 1 4 2 4 2 2
But it’s still not balanced, but we are 1 step closer!
It’s like a puzzle!

15. Tip 1: Only add 1 coefficient at a time. It is much easier this way
Tip 1: Only add 1 coefficient at a time. It is much easier this way! Tip 2: Always start by adding a coefficient to the more complex molecule. For example, chose H2O over O2. It will be easier to change O2 at the last minute, but it will be much harder if you wait till the end to manipulate H2O.

16. Step 3: ADD ANOTHER COEFFICIENT!
Now we need “4 H” on the reactants side (left) and “4 H” on the products side (right)
__ SnO2 + __ H2  __ Sn + __ H20 2 2 2 1
____ Sn ____ Sn
____ H ____ H
____ O ____ O
(reactants) (products) 1 1 4 4 4 2 2
We’ve done it! The equation is balanced!

17. Let’s try another example together! Step 1: Count the atoms
CH4 + O2  CO2 + H20
____ C ____ C
____ H ____ H
____ O ____ O
(reactants) (products) 1 1 2 4 2 s1 2 3 4
Uh ohhhhhhhh! The hydrogens AND the oxygens aren’t balanced!

18. (reactants) (products)
Step 2: Add coefficients Tip 2: It’s always a good idea to leave the single element, in this case O2 till last because it will be the easiest to change.
__ CH4 + __ O2  __ CO2 + __ H20 2
____ C ____ C
____ H ____ H
____ O ____ O
(reactants) (products) 1 1 2 4 4 s1 2 4 4
Now only the oxygens aren’t balanced!

19. (reactants) (products)
Step 3: Add a second coefficient Ask yourself: “How can I get 4 oxygens on the reactants side?”
__ CH4 + __ O2  __ CO2 + __ H20 2 2
____ C ____ C
____ H ____ H
____ O ____ O
(reactants) (products) 1 1 2 4 4 s1 4 4 4
Now we’ve balanced the chemical equation!

20. One more! Step 1: Count the atoms
SiO2 + HF  SiF4 + H20
____ Si ____ Si
____ H ____ H
____ O ____ O
____ F ____ F
(reactants) (products) 1 1 2 1 2 1 2 2 1 4 4
Uh ohhhhhhhh! The hydrogens AND the fluorines aren’t balanced!

21. __ SiO2 + __ HF  __ SiF4 + __ H20
Step 2: Add a coefficient! What atom is going to see the biggest increase? d F
__ SiO2 + __ HF  __ SiF4 + __ H20 4
____ Si ____ Si
____ H ____ H
____ O ____ O
____ F ____ F
(reactants) (products) 1 1 4 4 2 1 2 2 4 4 4
Now the fluorines are balanced!
But we still have an unequal amount of oxygens!

22. __ SiO2 + __ HF  __ SiF4 + __ H20
Step 3: Add another coefficient! Ask yourself: How can I get 4 oxygens on the products side?
__ SiO2 + __ HF  __ SiF4 + __ H20 4 2
____ Si ____ Si
____ H ____ H
____ O ____ O
____ F ____ F
(reactants) (products) 1 1 2 4 4 1 2 2 4 4 4
Now the equation is completely balanced!