Solving the Combustion of Octane

[tanya234]

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

 

[Clausius2]

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

Assuming a complete combustion:

$$C_8 H_{18} + a(O_2+3.714N_2)\rightarrow b CO_2 + c H_2O$$

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:

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

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.

 

[ravenprp]

To solve this problem, we need to use the balanced chemical equation for the combustion of octane and the given information about the composition and density of octane. Let's break down the steps to solve this problem:

Step 1: Calculate the molar mass of octane (C8H18) using the periodic table. The molar mass is 114.23 g/mol.

Step 2: Use the density of octane (0.70 g/mL) to calculate the mass of 1 litre of octane. 1 litre of octane is equal to 700 grams (0.70 g/mL x 1000 mL = 700 g).

Step 3: Calculate the number of moles of octane present in 700 grams using the molar mass. 700 g / 114.23 g/mol = 6.12 moles of octane.

Step 4: Use the mole ratio from the balanced chemical equation to determine the number of moles of oxygen needed for the combustion of 6.12 moles of octane. According to the equation, 1 mole of octane reacts with 25/2 moles of oxygen. Therefore, 6.12 moles of octane will react with (6.12 x 25/2) = 76.5 moles of oxygen.

Step 5: Use the given information about the composition of air to calculate the volume of air needed for 76.5 moles of oxygen. Since air is 22% oxygen by volume, we can assume that 22% of the volume of air is oxygen. Therefore, the total volume of air needed is (76.5 moles x 100%) / 22% = 347.7 moles. Now, we can use the conversion factor given in the note (1 mole of oxygen gas is equal to 24.5 litres) to calculate the volume of air needed in litres. (347.7 moles x 24.5 litres/mole) = 8,501.65 litres of air.

Therefore, 8,501.65 litres of air (78% N2, 22% O2 by volume) are needed for the combustion of 1 litre of octane. I hope this helps!