How would you write a balanced equation for the combustion of octane, C8H18 with oxygen to obtain carbon dioxide and water?

Answer 1

#2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O#

All you need to do is make sure that each element has the same number of atoms on both sides. You can do this step-by-step, beginning with the equation that is not balanced.

#C_8H_18 + O_2 -> CO_2 + H_2O#

By listing the elements and atom counts on both sides of the equation,

# 8-C-1# #18-H-2# #2-O-3#
Then as you can see, the molecule with 1 carbon (#CO_2#) you have to multiply by eight, the molecule with 2 hydrogens (#H_2O#) you have to multiply it by 9 to be equal to the left. And when you find that you have to multiply by a fraction then you can multiply the whole equation by an integer so you have only integers.
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Answer 2

The balanced chemical equation for the combustion of octane ((C_8H_{18})) with oxygen ((O_2)) to obtain carbon dioxide ((CO_2)) and water ((H_2O)) is:

[2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O]

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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.

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.

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.

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.

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