Question

(a) What is the total volume of gaseous products, measured at 350°C and 689 torr, when an automobile engine burns 103 g of C8H18 (a typical component of gasoline)? L (b) For part (a), the source of O2 is air, which is about 78.0% N2, 21.0% O2, and 1.0% Ar by volume. Assuming all the O2 reacts, but no N2 or Ar does, what is the total volume of gaseous exhaust?

Answer 1

Now as per the data provided we have C8H18 as the typical component of gasoline and O2 from air whose composition is provided in the question

since the question itself says 100 % conversion we will assume all of the O2 reacts to form CO2

Reaction:

Now

mass of C8H18 = 100 grams

molar mass of C8H18 = 114 grams/mol

Mols of C8H18 = 100/114 = 0.877 mol

From reaction Stoichiometry

Mols of O2 theoretically required = (25/2) mols of C8H18 = 12.5*0.877 = 10.9625 mols

Assuming 100% conversion

Mols of CO2 = 8* mols of C8H18 = 8*0.877 =7.016 mols

Mols of H20 = 9*0.877 =7.893 mols

now 21% of O2 is air

Air theoretically required =10.9625/0.21 =52.20234 mols

amount of nitrogen = 0.78*52.20234 =40.7178 mols

argon = 0.01*52.20234 =0.522 mols

Total mols = mols of argon + mols of nitrogen + mols of CO2 + mols of H20 = 0.522 + 40.7178 + 7.016+7.893 =96.617 mols

T = 350 oC = 623.15 K

P = 689 torr = 0.906 atm

n = 96.617 mols

By ideal gas law

PV = n R T

0.906* V =96.617 *0.0821*623.15

V = 5455.833 Liters