Balance the equation:-

C8H18+O2=CO2+H20

Balanced equation:
2 C₈H₁₈ + 25 O₂ = 16 CO₂ + 18 H₂O
Reaction type: combustion

Explanation:

First, you need to tally all the atoms.

C8H18(l) + O2(g) → CO2(g) + H2O(l)(unbalanced)

Based on the subscripts, you have

left side:
C = 8
H = 18
O = 2

right side:
C = 1
H = 2
O = 2 + 1 (do not add this up yet)

Second, find the easiest atom to balance. In this case, the C atom. Always remember that in balancing, you are NOT SUPPOSED TO CHANGE THE SUBSCRIPTS, only put coefficients before the substance (as changing the subscripts means that you are changing the molecular structure instead).

C8H18(l) + O2(g) → 8CO2(g) + H2O(l)

left side:
C = 8
H = 18
O = 2

right side:
C = (1 x 8) = 8
H = 2
O = (2 x 8) + 1

Since CO2 is a substance, you have to apply the coefficient to both C and two O atoms as they are all bonded to each other.

Third, balance the next easiest atom.

C8H18(l) + O2(g) → 8CO2(g) + 9H2O(l)

left side:
C = 8
H = 18
O = 2

right side:
C = (1 x 8) = 8
H = (2 x 9) = 18
O = (2 x 8) + (1 x 9) = 25

Now all that is left is to balance are the O atoms. Since the sum of O atoms on the right side is an odd number, I can use my knowledge in fractions to balance the left side of the equation.

Thus,

C8H18(l) + 25/2O2(g) → 8CO2(g) + 9H2O(l)(balance)

left side:
C = 8
H = 18
O = (2 x 25/2) = 25

right side:
C = (1 x 8) = 8
H = (2 x 9) = 18
O = (2 x 8) + (1 x 9) = 25

But if you don't want fractions as coefficients, you can always multiply the WHOLE equation by 2.

2 [C8H18(l) + 25/2O2(g) → 8CO2(g) + 9H2O(l)]

=

2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l) (balance)

Both answers are considered correct.