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DETAILED REPORT ON STOICHIOMETRIC RATIO CALCULATOR (WEEK-3 CHALLENGE) AIM To write a program that compares the effect of number carbon atoms on the stoichiometric co-efficient with respect to the combustion of different fuel type. EQUATIONS/ FORMULAE USED Generic alkane combustion, Cn H2n+2 + ar(O2 + 3.76 N2) = aCO2 +…
Ramkumar Venkatachalam
updated on 23 Mar 2021
DETAILED REPORT ON STOICHIOMETRIC RATIO CALCULATOR (WEEK-3 CHALLENGE)
To write a program that compares the effect of number carbon atoms on the stoichiometric co-efficient with respect to the combustion of different fuel type.
Generic alkane combustion, Cn H2n+2 + ar(O2 + 3.76 N2) = aCO2 + bH2O + cN2
Generic alkene combustion, Cn H2n + ar(O2 + 3.76 N2) = aCO2 + bH2O + cN2
Generic alkyne combustion, Cn H2n-2 + ar(O2 + 3.76 N2) = aCO2 + bH2O + cN2
Here
n = Number of atoms 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
Relation between stoichiometric co-efficient and number of atoms of carbon
Alkane combustion, ar = (3n+1)/2
Alkene combustion, ar = (3n)/2
Alkyne combustion, ar = (3n-1)/2
Programming language used – Octave 5.1.1
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Program
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% Stoichiometric cumbustion of Hydrocarbons
% n = number of atoms of carbon and a_r = stoichiometric coefficient
n = linspace(1,20,20);
% Alkane combustion
% Generic Alakne - Cn H2n+2
% Relation between n and ar =(3n+1)/2
a_r1 = (3*n+1)/2;
% Alkene combustion
% Generic Alkene - Cn H2n
% Relation between n and ar =3*n/2
a_r2 = (3*n)/2;
% Alkyne combustion
% Generic Alkyne - Cn H2n-2
% Relation between n and ar =(3n-1)/2
a_r3 = (3*n-1)/2;
% Plot comparing the relation between n and ar of Hydrocarbons
figure(1)
plot(n,a_r1, 'Linewidth',2)
hold on
plot(n,a_r2, 'Linewidth',2)
hold on
plot(n,a_r3, 'Linewidth',2)
axis([1 20 0 35])
title('Number of atoms of carbon vs Stoichiometric coefficient');
xlabel('Number of atoms of carbon','Fontsize',20);
ylabel('Stoichiometric coefficient','Fontsize',20);
legend('Alkane','Alkene','Alkyne');
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5. RESULTS
Graph
6. CONCLUSION
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