Oh my God, help please.I was absent so I didnt learn this and I dont get it at all. Help please!!!?

Question

For my science hw i have to balance equations. Here are some of the problems:

CH4 + O2 --> CO2 + H2O

Na + I2 --> NaI

N2 + O3 --> N2O

I have 12 other problems but I'm hoping i can figure them out after these. Can you explain to me in A LOT of detail how to balance them. I looked at so many websites and I'm even more confused now. Please help and dont just copy and paste something, explain it yourself. Also, I'm in 7th Grade Double XX but when you explain it pretend I'm like 10 years old. Thanks in advance.
<3bindorski

Answer 1

I've never been able to explain this stuff before, so sorry if I confuse you.

Okay, so when you balance equations, you want to make sure that on each side of the equation, each element has the same coefficient. What that means is that you want the number of atoms for each element to be equal.

So for
Na + I2 --> NaI,
that is not balanced because on the left side, there are 2 atoms of iodine. There are two Is, but on the right side, there is only one I.
What you want to do to make this equal is to add a 2 to the right side.

So Na + I2 ---> 2NaI
Unfortunately, the rule is that NaI is one unit, so you can't do anything like

Na + I2 ---> Na2I
or
Na + I2---> NaI2

So that's why you have to do it like
Na + I2 ---> 2NaI
But then you realize that there on the right side there are now 2 Nas.
So you have to balance that too.

2Na + I2 ---> 2NaI
Now you see that on both sides, there are two Nas and two Is.

Okay, now let's try it for N2 + O3 --> N20.
Notice how on the left side there are three Os, but on the right side there is only one. So what do you do?
You do this.
N2 + O3 --> 3N2O.
You see why I did that? So that there can be three Os on the right side. Unfortunately, there are now 6Ns on the right side and only 2 on the left side.
So simply add a three to the front of the N2 so that there are 6 nitrogens.
3N2 + O3 --> 3N2O.
So now you have 6 nitrogens and 3 oxygens on either side.

Let's try it for the hardest problem. (Oh, I'm kidding, it's not that hard.)

CH4 + O2 --> CO2 + H2O
Hmm...we now have three elements to worry about.
There is one C on each side.
There are 2 Os on the left side but 3 on the right.
There are 4 Hs on the left side but only 2 on the right.
Let's start with H. Starting with O will be harder, trust me.

CH4 + O2 --> CO2 + 2H2O
Do you see why I did that? It's so that there are four Hs on each side. There is one H4 and two H2's.
The Os are still unbalanced. There are still 2 on the left side, but now look at the right! Ta-da, there are four now. 2 in the CO2 and 2 in the H2O.

So we need to make sure there are four on the left, too.
CH4 + 2O2 --> CO2 + 2H2O.
Now count.
How many Cs? One on each side.
How many Hs? Four on each side.
How many Os? Four on each side.
The equation is balanced.

Answer 2

Reaction balancing is making sure that you have the elements on one side that you have on the other. You can't create them.

In the first problem, all the hydrogen comes from methane (CH4). If each water has two hydrogens, each CH4 can supply the hydrogen for two water molecules. So we can make a first change to:
CH4 + O2 -> CO2 + 2H2O
All the oxygen comes from O2 (oxygen exists in air as a dimer molecule), so for each CO2 formed, one O2 is needed. However, the oxygen
also is used in water formation. If we have two water molecules, we need one oxygen molecure to supply it. Now we can make a second change
CH4+ 2O2 -> CO2 + 2H2O

The second equation involves elemental iodine, which also is a dimer. So if we have I2 on one side, you need to have 2 iodides on the other.
The first change is
Na + I2 -> 2NaI
You need to double the sodium to balance:
2 Na + I2 -> 2 NaI

The third is a bit more difficult, but you should be able to work it out. Elemental nitrogen also exists as a dimer; here it reacts with OZONE, which is the trimer.

Answer 3

This is called stoicheometry. Total up the number of atoms on each side of the equation. Then, find the lowest commom denominator that will make each side equal. The first one:
left side C - 1, H - 4, O - 2
right side C 1, H - 2, O - 3
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2CH4 + 4O2 --> 2CO2 + 4H2O