Question

A vehicle burns gasoline (density = 0.79 kg/liter) at a rate of 7.6 liters per hour. Although gasoline is a mixture of many chemicals, for simplicity assume it has the formula of octane, C8H18.

 What is the required air flow rate in molar units (kmol/hr), in mass units (kg/hr) and in standard volume units (standard m3/hr) for air fed in stoichiometric proportion?

 Determine the flowrate (kmol/hr) and composition (mol fraction) of the exhaust gases if 25% excess air is fed with the gasoline. For purposes of this problem, only carbon dioxide and water (no carbon monoxide) is produced in the reaction and all of the gasoline is consumed.

Answer 1

Balanced chemical reaction is as follow

2 C₈H₁₈ + 25O₂ 16 CO₂ +18 H₂O

density of octane is 0.79 kg/liter then 7.6 liter contain 0.797.6 = 6.004 kg octane = 6004 gm octane

molar mass of octane = 114.23 g/mol

6004/114.23 = mole of octane

6004 gm octane = 52.5606 mole of octane

52.5606 mole of octane burn per hour

According to reaction for burning 2 mole of octane require 25 mole of Oxygen

then 52.5606 mole of octane require 2552.5606/2 = 657.0075 mole of oxygen

required air flow rate in molar units = 0.657kmol/hr

1 mole O₂ = 31.9988 gm then 657.0075 mole O₂ = 657.0075 31.9988 = 21023.45 gm

required air flow rate in mass unit = 21.02345kg/hr

According to avogadro's law 1 mole of O₂ = 0.022414m³ volume then 657.0075 mole O₂ =657.00750.022414 =

14.726 m³

required air flow rate in standard volume unit = 14.726 m³/hr

2 mole of octane produce 16 mole of CO₂ then 52.5606 mole octane produce 52.560616/2 = 420.4848 mole of CO₂

flowrate of CO₂ = 0.4204848kmol/hr

2 mole of octane produce 18 mole of H₂O then 52.5606 mole octane produce 52.560618/2 = 473.0454 mole of H₂O

flowrate of H₂O = 0.4730454kmol/hr