Gas Laws Chemistry Question!!

Octane, C8H18, is one of the hydrocarbons in petrol. Upon combustion octane produces CO2 and H2O. What volume of oxygen measured at 108.5 kPa and 27.5oC is required to burn 1.80 g octane? (Take MW octane as 114)

a. 0.364 L

b. 4.55 L

c. 41.4 L

d. 416 mL

e. 44.9 mL

You need a balanced equation.

C8H18 + 25/2 O2 ---------------> 8CO2 + 9HOH

Convert the 1.8 g of octane into moles

1.8 / 114 = 0.015789 mol

Now the equation says that you need 25/2 more moles of O2 as you have of octanne. So moles of O2 needed will be

0.015789 mol x 25/2 or 0.197368 moles

Now use the Ideal Gas Law to solve for volume

V = nRT/ P

V = (0.197368 x 8.31 x 300.5) / 108.5 = 4.55 L