How can I balance an equation with odd numbers of certain elements?

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Balancing chemical equations with odd numbers of elements can be challenging, particularly with hydrocarbons like C2H6. The key is to first balance the carbon and hydrogen atoms before addressing oxygen. In the example provided, starting with one molecule of C2H6 leads to the production of two CO2 and three H2O, which requires calculating the oxygen needed. This results in a fractional coefficient for O2, specifically 3.5, which can be eliminated by multiplying all coefficients by 2 for a balanced equation. Understanding this method simplifies the process of balancing equations with odd numbers of certain elements.
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I know this is elementary stuff, but I'm in Gen Chem 1 and this is the only thing I am having trouble with. I am fine balancing most equations, but when there are odd numbers of certain elements I get sort of confused. For example:

C2H6 + O2 \rightarrow CO2 + H2O


The Attempt at a Solution



I have tried several times, but the closest I could get was:

2C2H6 + 5O2 \rightarrow 2CO2 + 6H2O

It is the Oxygen andHydrogen being uneven that is screwing me up. Right away I realize I have 3 Oxygen on the right and 2 on the left, so the first thing I tried to do was make both sides have 6 oxygen...

C2H6 + 3O2 \rightarrow CO2 + 4H2O

But then that leaves me with 8 Hydrogen on the right, and only 6 on the left...I can't think of any ways to get that 6 to equal 8! Then I tried for 12 Oxygen on both sides instead of 6...then my problem is still Hydrogen.

So maybe y'all see what I'm confused about...like I said, this is the only thing I am having trouble with so far in Gen Chem is balancing this specific type of equation, and we have our first test on Tuesday! So hopefully someone can help :)
 
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Let's start with 1 molecule of C2H6. It is obvious we will produce two CO2 and three H2O:

C2H6 + xO2 → 2CO2 + 3H2O

Can you find x? It will be a fraction now - but you can multiply everything by some small integer to get rid of the denominator...
 
It would be 3\frac{1}{2}, yes? So I should then multiply by a small integer to get rid of denom, multiply all by 2? Does that include the 3.5?
 
Yes, multiply 3.5 as well.
 
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