Question

Assume that a single cylinder of an automobile engine has a volume of 523 cm3 .

a. If the cylinder is full of air at 77 ∘C and 0.980 atm , how many moles of O2 are present? (The mole fraction of O2 in dry air is 0.2095.)

b. How many grams of C8H18 could be combusted by this quantity of O2, assuming complete combustion with formation of CO2 and H2O?

Answer 1

(a) moles of air can be calculated as

PV = nRT

P = 0.98atm, V = 523cm3 = 0.0523L, R = 0.08206 Latm/Lmol, T = 273 + 77 =350K, n = moles of air

0.98 X 0.523 = n X 0.08206 X 350

n = 0.98 X 0.523 / 0.08206 X 350 = 0.017

Mole fraction of O2 = moles of O2 / total moles of Air

moles of O2 = Mole fraction of O2 X total moles of Air = 0.2095 X 0.017 = 0.00356

(b) Write the balanced equation

C8H18 + 25/2O2 -----------> 8CO2 + 9H2O

one mole of C8H18 or 114g require 25/2 or 25/2 X 32 = 400g of O2

Because molecular mass of C8H18 = 114g and Mol mass of O2 = 32g

mass of O2 in grams present in the automobile engine = mol. mass X no. moles

mass of O2 in grams present in the automobile engine = 32 X 0.00356 = 0.11392g

400g of O2 = 114g of C8H18

Or 1 g of O2 = 114 / 400 = 0.285g

Above result means 1g of O2 will be needed for the combustion of 0.285g of C8H18

Therefore 0.11392g of O2 will combust 0.285 X 0.11392 = = 0.0325g

Therefore 0.0325g of C8H18 will be combusted by this quantity of O2