Question

A chemical analysis of a sample of ordinary diesel fuel reveals that it is composed of 86% carbon (by mass) and 14% hydrogen (by mass)

What is the air-fuel ratio on a mass basis for this stoichiometric reaction?

Suppose the fuel is burned with 175% theoretical air; what is the reaction’s equivalence ratio?

Answer 1

C = 86% if C

H = 14%

find molar ratio

mol of C = mass/MW = 86/12 = 7.16666

mol of H = mass/MW = 14/1 = 14

ratio is:

14/7.166 = 2 approx...

so

CH2

then

CH2 + O2 = CO2 + H2O

ratio is:

1 mol of fuel : 1 mol of O2

so

MW of fuel = 12*1 + 1*2 = 14 g/mol

mol = mass/MW = 100/14 = 7.1428 mol

so

7.1428 mol of fuel --> 7.1428 mol of O2

mass of fuel = 100 g of fuel

mass of air = 7.1428/0.21 * 29 = 986.3 g of air

fo ratio

986.3 /100 = 9.863 ratio

so..

for 175% air...

986.3 g of air = 100% so for 175% --> 175/100*986.3 = 1726.025 g of air requried

so new ratio

1726.025 / 100 = 17.2602