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Perpetual Basic Chemistry 101 Question Thread. Plz Help!

KAMAZON

KAMAZON

Golden Member
Hello, I had this chemistry question that really taking a lot of my time and I can't really ask the teacher for too much help since it's kind of hard to really understand him. I wanted to see if anyone can help me out with this example:

http://www.brazosport.cc.tx.us/~chem/tutor/baleqn.html


Example 2:
Balance the following equation which shows the combustion (or burning) of propane:

C3H8 + O2 => CO2 + H2O

Solution 2:
There are 3 carbons on the left, but only 1 on the right. There are 8 hydrogens on the left, but only 2 on the right. There are 2 oxygens on the left, but 3 on the right.

Of the three atoms, carbon and hydrogen are the best to start with. We should save oxygen for last for two reasons. It appears by itself as the element on the left, and it appears in more than one place on the right.

Starting with carbon, we can put a 3 in front of CO2 on the right side to balance out the carbons. Likewise, we can put a 4 in front of H2O on the right side to balance the hydrogens.

C3H8 + O2 => 3 CO2 + 4 H2O

This gives us a total of 3 carbons on both sides, 8 hydrogens on both sides, 2 oxygens on the left and 10 oxygens on the right. To balance the oxygens, we put a 5 in front of O2 on the left side.


Ok i can see that you have to add 4 Hydrogen, but why does the 4 NOT multiply into the oxygen? Isn't the distribution property supposed to be used on this and if not, why not? Any clarification is greatly appreciated. Thanks.

 
I'm confused by your question. The 4 does multiple the # of oxygen.

the final solution should be C3H8 + 5 O2 => 3 CO2 + 4 H2O

 
You nailed my question right on the head lol. I'm wondering WHY the 4 does not apply to the oxygen. It seems like if you're going to multiply the Hydrogen by 4, then why wouldn't you also have to apply that to the Oxygen since it is part of the same compound?

 
Originally posted by: KAMAZON
You nailed my question right on the head lol. I'm wondering WHY the 4 does not apply to the oxygen. It seems like if you're going to multiply the Hydrogen by 4, then why wouldn't you also have to apply that to the Oxygen since it is part of the same compound?
Click to expand...

I apologize, this is a simple problem so I still don't see your confusion.

The 4 does multiple to the oxygen. You have 4 molecules of H2 which 8 total hydrogen atoms and 4 oxygen atoms.

Are you hung up on the explaination given maybe? Again I'm sorry, I'm not seeing your trouble.

My way of doing this would be you comparing each element one at a time from to right.

First balance, the carbon, then hydrogen, then oxygen.

that's exactly how the example does it too.
 
What's the problem.. the 4 does multiply into the oxygen.

Left Side
5O2 = 5x2 = 10 Oxygen

Right Side
3CO2+4H2O=3x2+4x1=10 Oxygen
 
Originally posted by: KAMAZON
Hello, I had this chemistry question that really taking a lot of my time and I can't really ask the teacher for too much help since it's kind of hard to really understand him. I wanted to see if anyone can help me out with this example:

http://www.brazosport.cc.tx.us/~chem/tutor/baleqn.html


Example 2:
Balance the following equation which shows the combustion (or burning) of propane:

C3H8 + O2 => CO2 + H2O

Solution 2:
There are 3 carbons on the left, but only 1 on the right. There are 8 hydrogens on the left, but only 2 on the right. There are 2 oxygens on the left, but 3 on the right.

Of the three atoms, carbon and hydrogen are the best to start with. We should save oxygen for last for two reasons. It appears by itself as the element on the left, and it appears in more than one place on the right.

Starting with carbon, we can put a 3 in front of CO2 on the right side to balance out the carbons. Likewise, we can put a 4 in front of H2O on the right side to balance the hydrogens.

C3H8 + O2 => 3 CO2 + 4 H2O

This gives us a total of 3 carbons on both sides, 8 hydrogens on both sides, 2 oxygens on the left and 10 oxygens on the right. To balance the oxygens, we put a 5 in front of O2 on the left side.


Ok i can see that you have to add 4 Hydrogen, but why does the 4 NOT multiply into the oxygen? Isn't the distribution property supposed to be used on this and if not, why not? Any clarification is greatly appreciated. Thanks.
Click to expand...

It does. There should be a coefficient of 5 infront of the O2 on the left, but it looks as if due to a type (either from the book or you) its missing. Its stated in the description in the bolded part below.

This gives us a total of 3 carbons on both sides, 8 hydrogens on both sides, 2 oxygens on the left and 10 oxygens on the right. To balance the oxygens, we put a 5 in front of O2 on the left side.
 
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