The equation for the complete combustion of octane is as follows

2C8H18 +25O2—–16CO2+H20

When 228g of octane are burnt completey.

Calculate 1. the volume of oxygen needed

2. the volume of carbon dioxide produced

(Assume that 1 mole of gas has a volume of 22.4dm3)

Hi!!

The equation gives the moles ratio between reactants and products. For

case, to burn 2 moles of C8H18, 25 moles of O2 are needed, 16 moles of

CO2 and 1 single mole of water are produced.

1 mole of octane weights (8*12g + 18*1g) = 114g

So 228g of octane are equal to (228/114) = 2 moles of octane; then if

the 228g of octane are burnt completely; this means that 2 moles of

octane are burnt completely; therefore 25 moles of O2 are needed.

Since each mole has a volume of 22.4dm3, the volume of oxygen needed

is (25*22.4dm3) = 560dm3

As a product of the reaction 16 moles of CO2 are produced, this amount

of CO2 has a volume of (16*22.4dm3) = 358.4dm3

I hope this helps you.

Regards,

livioflores-ga

Hi!!

The commenter momomc-ga is right, the balanced equation is:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Affortunately the only part that is wrong in the original misbalanced

equation was the "+ 18H2O", specifically the number of moles of water

that the combustion produces. Since the other terms of the equation

are right and no question was made about the water, the answer does

not change.

Regards,

livioflores-ga