Octane, a component of gasoline, reacts with O2 according to the equation: 2 C8H

Question

Octane, a component of gasoline, reacts with O2 according to the equation: 2 C8H18(g) + 25 O2(g) --> 16 CO2 (g) + 18 H2O (g). assume the volume of one cylinder of an automobile engine is 500 cm3. If air enters the engine at 50 degrees Celsius and 1.0 atm of pressure how many grams of octane should the fuel injection system send to the cylinder in order to completely consume all of the O2 in the air? assume that the mole fraction of O2 in the air is 0.20. 2. Octane, a component of gasoline, reacts with O2 according to the equation: 16C0218 H20 (e Assume the volume of one cylinder of an automobile engine is 500 cm3. If air enters the engine at 50°C and 1.0 atm of pressure, how many grams of octane should the fuel injection system send to the cylinder in order to completely consume all of the O2 in the air? Assume that the mole fraction of O2 in the air is 2C8Hj 8 (g) 2502 (g) > 16 CO2 (g) 18 H20 (g) + - 0.20

Explanation / Answer

Given : Volume of cylinder = 500 cm3 . T = 50 deg C = 50+273.15 = 323.15 K

Pressure = 1.0 atm

Mole fraction of O2 in the air is 0.20

Solution: we have to calculate number of moles of O2 by using mole fraction.

We calculate total moles in the 500 cm3 = 0.500 L

Number of moles of air can be calculated by using ideal gas law

pV = nRT

n = pV / RT

p pressure in atm , V volume in L , R is gas constant value is 0.08206 L atm per K per mol

T is temperature in K

n (air) = 1 atm * 0.500 L / (0.08206 L atm / (K mol) * 323.15 K )

= 0.01885 mol

By using mol fraction we find moles of O2 in air

Moles of air = 0.01885 mol air * 0.2 = 0.003771 mol O2

Now we calculate moles of octane required to burn with 0.003771 moles of O2

Moles of octane = moles of oxygen * 2 mol octane / 25 moles of O2

( mol ratio of octane to O2 is 2 : 25)

Moles of octane = 0.003771 moles of oxygen * 2 mol octane / 25 moles of O2

= 0.0003017 mol octane

Mass of octane is calculated by using molar mass and moles of octane

Mass of octane in g = 0.0003017 mol octane * 114.2302 g per mol

= 0.03446 g octane

Mass of octane = 0.03446 g

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