can someone help me balance these chemical equations???

Question

I am not understanding the process of balancing equations. Can someone please help me???


1.) Na(subscript 2)O(subscript 2) + H(subscript 2)O *arrow* NaOH + O(subscript 2)

2.) H(subscript 3)BO(subscript 3) *arrow* H(subscript 4)B(subscript 6)O(subscript 11) + H(subscript 2)O

3.) FeS(subscript 2) + O(subscript 2) *arrow* Fe(subscript 2)O(subscript 3) + SO(subscript 2)

4.) C(subscript 3)H(subscript 8) + O(subscript 2) *arrow* CO(subscript 2) + H(subscript 2)O

Answer 1

These are pretty easy, as is the whole process.

The trick is to remember that the number of atoms of all types must be equal on both sides of the equation -- just like in math.

Let's start with the first one:

1) w Na2O2 + x H2O ===> y NaOH + z O2

you have 2 sodium (Na), two hydrogen (H) and three oxygen on the left, so that's how many you should end up with on the right. In order to get two sodiums on the right, y = 2, so

Na2O2 + x H2O =====> 2 NaOH + z O2

now, if you count up the right side, you have two sodium, two hydrogen, and 2 + 2z oxygen atoms. How can I make 2 + 2z = 3? If I let z = 1/2, the equation balances!

Na2O2 + H2O ====> 2 NaOH = ½O2

Now, try that same process on trhe rest of the examples from your homework . . .

2) x H3BO3 ====> y H4B6O11 + z H2O

3) w FeS2 + x O2 ====> y Fe2O3 + z SO2

4) w C3H8 + x O2 ====> y CO2 + z H2O
(the combustion of octane)
just for a check on your progress, in this one your factors should be w=1, x= 5, y=3, and z=4. Did you match my answer?

Answer 2

1) Na2O2 + H2O --> NaOH + O2
This one is very simple.
Basically, count how many Na's there are on each side.
There's two on the left and one on the right.
Put a 2 in front of NaOH.
But now, you have 2 more oxygens and 2 more hydrogens
So you put a 2 in front of H2O

Na2O2 + 2H2O ---> 2NaOH + O2

The equation is now balanced with 2 sodiums, 6 oxygens, and 2 hydrogens on each side.

2) H3BO3 --> H4B6O11 + H2O

First, you need to basically mess around for a little bit with this one. Try random coefficients for H2O - 7, even though an odd number, works here because 11 + 7 = 18, an even number of oxygens.
Now because you have 18 oxygens on the right side, you need to make it 18 on the left side. Put the coefficient of 6 for H3BO3.

Done! lol

6H3BO3 --> H4B6O11 + 7H2O

3) FeS2 + O2 --> Fe2O3 + SO2

This is an easier equation.

Because there's an odd number of oxygens on the right side, you should put a 2 as the coefficient for Fe2O3. [You almost ALWAYS want to make any odd number of elements even to balance equation)

Now because of the new coefficient, there are 4 Fe's. Put a 4 as the coeff. for FeS2.

Now you have 8 S's on the left side, so you put an 8 as the coeff. for SO2.

And finally now you have 22 oxygens on the right side of the equation - so put an 11 as the coeff. for O2.

4FeS2 + 11O2 --> 2Fe2O3 + 8SO2 = answer

4) C3H8 + O2 --> CO2 + H2O

Here you should first make the C3H8 even - put 2 as a coefficient.
This causes 16 H's to be on the left side. Put an 8 as the coeff. for H2O to balance it.
This also causes 6 Carbons to be on the left side - put a 6 as a coeff for CO2.
Now, you have 20 oxygens on the right side after doing the previous steps - put a 10 in front of the O2.

Now you should get this:
2C3H8 + 10O2 --> 6CO2 + 8H2O
which can be simplified/divided by 2 to make..

C3H8 + 5O2 --> 3CO2 + 4H2O

Answer 3

If you count out esch one seperately then just make sure the numbers are the same on each side. Because matter is never created or destroyed in a reaction.

2Na2O2 + 2H2O ---> 4NaOH + O2
Na- 4 Na-4
O-6 O - 6
H-4 H - 4

6H3BO3---> H4B6O11 + 7H2O
H-18 H-18
B-6 B-6
O-18 O-18

4FeS2 + 11O2---> 2Fe2O3 + 8SO2
Fe-4 Fe-4
S-8 S-8
O-22 O-22

C3H8 + 5O2---> 3CO2 + 4H2O
C-3 C-3
H-8 H-8
O-10 O-10