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How do you balance #C_2H_6 + O_2 -> CH_3COOH + H_2O#?

Answer 1

Caleb Adams

2C₂H₆ + 7O₂ -> 4CH₃COOH + 6H₂O

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Answer 2

Owen Ambrose

#C_2H_6 + 3/2O_2 rarr CH_3COOH + H_2O# or #2C_2H_6 + 3O_2 rarr 2CH_3COOH + 2H_2O#

Count the number of atoms of each element on both sides of the reaction arrow by performing an atom inventory first:

Side of the reactants: C atoms = 2 H atoms = 6 O atoms = 2

H atoms = 6 O atoms = 3 C atoms on the products side

As we can see the only element that isn't balanced is oxygen. A simple approach to this would be to focus on the #O_2# on the reactants side. You want to find a coefficient that would make the number of oxygen atoms on both side to be the same.

Place a coefficient of 3/2 in front of #O_2# so the total number of atoms will equal three since you always multiply the coefficient by the subscript that's attached to that atom.

Then you'll end up with this #C_2H_6 + 3/2O_2 rarr CH_3COOH + H_2O#

If fractional coefficients bother you, you can also obtain whole number coefficients by multiplying the entire chemical reaction by two, as in this example:

#2C_2H_6 + 3O_2 rarr 2CH_3COOH + 2H_2O#

I hope this is clear.

Help with Balancing Equations

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you balance #C_2H_6 + O_2 -&gt; CH_3COOH + H_2O#?

How do you balance #C_2H_6 + O_2 -> CH_3COOH + H_2O#?