How many liters of air (78 percent N2, 22 percent O2 by volume) at 20 degrees Celsius and 1.00 atm are needed for the complete combustion of 1.0 L of octane, C8H18, a typical gasoline component that has a density of 0.70 g/mL?
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To find the number of liters of air needed for the complete combustion of octane, we need to consider the stoichiometry of the combustion reaction and the balanced chemical equation.
The balanced chemical equation for the combustion of octane (C8H18) is:
C8H18 + 12.5O2 -> 8CO2 + 9H2O
From the balanced equation, we can see that for every 1 mole of octane, we need 12.5 moles of oxygen (O2) to have a complete combustion.
First, let's determine the number of moles of octane in 1.0 L of octane. Since the density of octane is given as 0.70 g/mL, we can calculate the mass of 1.0 L of octane:
Mass = density * volume = 0.70 g/mL * 1000 mL = 700 g
Next, we need to convert the mass of octane to moles using its molar mass. The molar mass of octane (C8H18) can be calculated by adding up the atomic masses of its constituents:
Molar mass of C8H18 = (8 * atomic mass of C) + (18 * atomic mass of H)
= (8 * 12.01 g/mol) + (18 * 1.01 g/mol)
= 114.23 g/mol
Moles of octane = mass / molar mass = 700 g / 114.23 g/mol ≈ 6.12 mol
Now, using the stoichiometry from the balanced equation, we know that for every 6.12 moles of octane, we need 12.5 moles of O2. Since air is 22% O2 by volume, we can determine the volume of air in liters that contains 12.5 moles of O2:
Volume of O2 = (12.5 mol O2) * (22% volume O2 / 100%)
= 2.75 mol O2
To convert moles of O2 to liters of air, we need to use the ideal gas law equation:
PV = nRT
Assuming the temperature (T) is 20 degrees Celsius, which is 293 Kelvin, and the pressure (P) is 1.00 atm, we can solve for the volume (V) in liters:
V = nRT / P
= (2.75 mol) * (0.0821 L*atm/(mol*K)) * 293 K / 1.00 atm
≈ 66.91 L
Therefore, approximately 66.91 liters of air (at 20 degrees Celsius and 1.00 atm) are needed for the complete combustion of 1.0 L of octane.
(1) Compute the number of moles (n) of octane in 1 liter. The mass of that liter is 700 g, because of the 0.70 g/ml density.
(2) Using the balanced reaction for oxidation, calculate how many moles of O2 are needed to burn n moles of C8H18. That number is 12.5.
C8H18 + 12.5 O2 = 8CO2 + 9H2O
(3) The number of moles of air needed will be 12.5 n/0.22 = 56.8 n
(4) One mole of any gas occupies 24.2 liters at 20 C and 1 atm.
Put the pieces together for the final answer.