Could you please show me the steps of balancing that chemical equation?

Balancing Equation is Kind of Complicated, But What u have to do is just make sure that the right hand Side Equals the Left hand side.

Here on the left hand side .. there is h20 ... In Chemical Equation ... u can never have a ½ h20 .. So there lies the Key in this Equation

But Also Remember U cant have ½ 02 molecule either

So when u try to balance the equation .. to get an even number of h20.. the least number od c5h90 u have to have is 2 ... but when u try yo balance the equation... U will have 13½ O2 Molecule therefore, u have to try 4 which will give even number of H2O and it will balance the equation by fulfilling RHS = LHS

4 C5H9O + 27 O2 ------> 20 CO2 + 18 H2O

_____C5H9O + _____O2 ____CO2 + _____H2O

For starters, balance C5 and 5 CO2

_____C5H9O + _____O2 ____5CO2 + _____H2O

Next, 9H won't work--we need an even number, so double the amount

2C5H9O + O2---> 10CO2 + 9H2O

Now, can we get the Os balanced? No, we need an even number of O2 to make it work so double it again

4C5H9O + 27O2---> 20CO2 + 18H2O

First count the moles of each element on both sides:

Left
C = 5
H = 9
O = 1 + 2 = 3

Right
C = 1
H = 2
O = 2 + 1 = 3

C and H are not balance therefore we go from there.  Obviously, since oxygen are in all the compounds, changing anything will also change oxygen counts.  However, we won't be worrying about oxygen since there's oxygen alone on the left.

Start modifying on the smaller counts of element such as H or C on the right side.  For no specifically reason, I'll start balancing carbon on the right side by making 5 moles of CO2, this will give:

___ C5H9O + ___ O2 5 CO2 + ___ H2O

Left:
C = 5
H = 9
O = 3

Right:
C = 5 x 1 = 5
H = 2
O = 5 x 2 + 1 = 11

C is now balanced, we'll start with H next.  Notice however that H on the left side is an odd number and H on the right side is even, this will be important later.  But right now, just modify so that you have 9/2 moles of H2O:

___ C5H9O + ___ O2 5 CO2 + 9/2 H2O

Left:
C = 5
H = 9
O = 3

Right:
C = 5 x 1 = 5
H = 9/2 x 2 = 9
O = 5 x 2 + 9/2 x 1 = 14.5

Now that C and H are balanced, that left with just oxygen.  But we got oxygen on the left side alone, we can change just that.  6.75 (27/4) moles of O2 will be enough to balance the entire equation.

1 C5H9O + 27/4 O2 5 CO2 + 9/2 H2O

Left:
C = 5
H = 9
O = 1 + 27/4 x 2 = 1 + 27/2 = 14.5

Right:
C  = 5 x 1 = 5
H = 9/2 x 2 = 9
O = 5 x 2 + 9/2 x 1 = 14.5

But since fractions are not proper form, you multiply everything by 4 to get rid of the fractions:

4 C5H9O + 27 O2 20 CO2 + 18 H2O