Mole question

  • Thread starter tanya234
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Hello, how would i solve the following?:

"How many litres of air (78% N2, 22% O2 by volume) are needed for the combustion of 1 litre of octane, C8H18, a typical gasoline component, of density 0.70g/mL?"

Note - 1 mole of oxygen gas is equal to 24.5 litres.


I have done the chemical equation of the combustion of octane which is C8H18 + 25/2O2 ---> 8CO2 + 9H20 but I am unsure about what to do next.

Thank you :smile:
 

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  • #2
Clausius2
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tanya234 said:
Hello, how would i solve the following?:

"How many litres of air (78% N2, 22% O2 by volume) are needed for the combustion of 1 litre of octane, C8H18, a typical gasoline component, of density 0.70g/mL?"

Note - 1 mole of oxygen gas is equal to 24.5 litres.


I have done the chemical equation of the combustion of octane which is C8H18 + 25/2O2 ---> 8CO2 + 9H20 but I am unsure about what to do next.

Thank you :smile:
Assuming a complete combustion:

[tex] C_8 H_{18} + a(O_2+3.714N_2)\rightarrow b CO_2 + c H_2O[/tex]

Conservation of atoms:

b=8, c=9, a=25/2 (you were right)

So that, the Fuel to Air ratio at stochiometric conditions is: W=molecular weight, N=moles, m=mass, f=fuel, a=air:

[tex] FAR_s=\frac{m_f}{m_a}=\frac{W_f N_f}{W_a N_a}=\frac{W_f}{4.714W_a a}=\frac{114 g/mol}{4.714\cdot 28.9 \cdot 25/2}=\frac{0.0669 g_{fuel}}{g_{air}}[/tex]

1 litre of octane is 0.7 Kg. So that you will need 0.7/0.066=10.4 Kg of air, which at normal conditions has a density of 0.0012 Kg per litre, therefore 8.6 cubic meters of air will be needed.
 

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