Stoichiometric Co-efficient Evaluation

Fuel Types: -

1) Alkane - Alkane is the type of acrylic based structured hydrocarbon, in which the carbon and hydrogen atoms are arranged in the tree like structure. The alkane have general formula of CnH2n+2">CnH2n+2CnH2n+2.

2) Alkene - It is an unsaturated hydrocarbon which contain carbon - carbon double bond, the general formula for the alkene is CnH2n">CnH2nCnH2n. It has the two lover hydrogen atom corresponding to the Alkane of same carbon atoms.

3) Alkyne - It is similar to above given fuel type the difference is just it has atleast one carbon - carbon triple bond. The chemical formular for the alkyne is CnHn-2">CnHn−2CnHn-2.

The task is to calculate the stoichiometric coefficient for given chemical reaction of Alkane fuel. It is important to know what is the stoichiometric coefficient which will give the idea how to calculate from the reactant and product side of reaction.

Stoichiometric Co-efficient: - It is coefficient written in front of atom or ion in the chemical reaction to balance the number of each element or molecules in both reactant and product side of reaction. The stoichiometric ratio is important in the field of chemistry or combustion science to determine number of moles required for the completion of chemical reaction.

The fuel combustion process equation is given as:

CnH2n+2+ar(O2+3.76N2)=aCO2+bH2O+cN2">CnH2n+2+ar(O2+3.76N2)=aCO2+bH2O+cN2CnH2n+2+ar(O2+3.76N2)=aCO2+bH2O+cN2

Where,

n = Number of moles of C

ar = Stoichiometric co-efficient

a = Number of moles of Carbon-di-oxide

b = Number of moles of water

c = Number of moles of nitrogen

To calculate the 'ar' in each of the fuel type it require to compare the each element of reactant and product in the reaction.

Consider Alkane equation and carbon element in LHS and RHS of equation,

by calculating we get and equation as, n=a

similarly for Hydrogen element the equation we get is, b = n+1

from oxygen element we get the equation for ar which is,

ar=3n+12">ar=3n+12ar=3n+12

So according to above equation it is relating stoichiometric coefficient and number of carbon. This help us to plot the graph accordingly for all fuel types.

The code for calculating the "ar" is given below.

clear all
close all
clc
%For Alkane fuel
n = [1:10];   % Number of moles Line 1
HC1 = (2*n+2); %hydrocarbon chain Line 2
a = n; % substituting number of mole to carbon-di-oxide  Line 3
b1 = HC1/2; % calculating number of moles for water element Line 4
ar1 = (2*a+b1)/2; %calculating stoichiometric coefficient for Alkane Line 5

%For Alkene fuel type

HC2 = (2*n);
a1 = n;
b2 = HC2/2;
ar2 = (2*a1+b2)/2;

%For Alkyne fuel type

HC3 = (2*n-2);
a2 = n;
b3 = HC3/2;
ar3 = (2*a2+b3)/2;

%Plotting of Results

plot(n,ar1,'r','linewidth', 3)
hold on
plot(n,ar2,'b','linewidth', 3)
hold on
plot(n,ar3,'y','linewidth', 3)
xlabel('Number of Moles of Carbon')
ylabel('Stoichiometric Coefficients')
legend('ALKANE','ALKENE','ALKYNE')

So in the code as we can see above:

Line 1 which represent the number of moles for which stoichiometric coefficient is get calculated. After that the chemical formula for Alkane in which only hydrocarbon chain is used to calculate the ar is used with moles of water molecules as we can see in the line number 5.

Results: -

The results are plotted for Alkane, Alkene and Alkyne fuel type.

From the above graph it is seen that, the stoichiometric coefficient is linear with number of moles of carbon. For a specific number of moles of carbon the stoichiometric coefficient is higher in the Alkane and least in the Alkyne.

Conclusion: -

1) Alkane require more amount of oxygen moles to burn the one mole of carbon, as there is more oxygen to burn the fule the by-product will be burning completely leaving no harmful reaction.

2) On the other hand as Alkyne require less number of mole of oxygen to burn the one mole of carbon. As the Alkyne are triple bond unsaturated hydrocarbon while burning it will mostly create and incomplete combustion with higher energy factor.