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In the theory of the Stoichiometric combustion assuming when the fuel and air mix, the products produced are only CO2, H2O, and N2. …
ARTH SOJITRA
updated on 27 May 2020
In the theory of the Stoichiometric combustion assuming when the fuel and air mix, the products produced are only CO2, H2O, and N2.
ALKANES
The governing equation for the stoichiometric combustion for alkane is :
CnH2n+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
Assuming 100% yeild rate ; we get :
n = a;
2n +2 = 2b or n + 1 = b;
2ar = 2a + b or 2ar = 2(n) + ( n + 1 ) or ar = 3n+12; and
3.76 ar = c or 3.76×(3n+1)2 = c = 5.64n + 1.88 ;
In the problem statement, it is given to compare the relationship between the stoichiometric co-efficient ar and the number of moles of carbon i.e. n
From the above equations we have,
ar = 3n+12
In the plot between ar vs. n, we will have a graph with a positive slope of +1.5 and varying linearly.
ALKENES
The governing equation for the stoichiometric combustion for an alkene is :
CnH2n + 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
Assuming 100% yeild rate ; we get :
n = a;
2n = 2b or n = b;
2ar = 2a + b or 2ar = 2(n) + (n) = 3n2; and
3.76 ar = c or 3.76×3n2 = c = 5.64n ;
In the problem statement, it is given to compare the relationship between the stoichiometric co-efficient ar and the number of moles of carbon i.e. n
From the above equations we have,
ar = 3n2
In the plot between ar vs. n, we will have a graph with a positive slope of +1.5 and varying linearly.
ALKYNES
The governing equation for the stoichiometric combustion for an alkyne is :
CnH2n-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
Assuming 100% yeild rate ; we get :
n = a;
2n - 2 = 2b or n - 1 = b;
2ar = 2a + b or 2ar = 2(n) + (n - 1) = 3n−12; and
3.76 ar = c or 3.76×(3n−1)2 = c = 5.64n - 1.88 ;
In the problem statement, it is given to compare the relationship between the stoichiometric co-efficient ar and the number of moles of carbon i.e. n
From the above equations we have,
ar = 3n−12
In the plot between ar vs. n, we will have a graph with a positive slope of +1.5 and varying linearly.
Conclusion: The variation of the number of moles of carbon on the stoichiometric coefficient is linear as is confirmed in the MATLAB study.
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