Question In Description

Question

B) If a cylinder in an automobile engine has a volume of 525 cm3, how many moles of oxygen would be present at 90.0°C and 0.995 atm? Air is 20.95 percent oxygen

C) Calculate the maximum number of grams of octane that the oxygen in part B) could completely burn

Answer 1

B) Ideal gas law: PV = nRT

P = pressure = 0.995 atm

V = volume in liters = 525 cm³ x 1 L / 1000 cm³ = 0.525 L

n = moles of air =?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 90.0ºC + 273.15 = 363.15K

Solve for n:

n = PV/RT = (0.995)(0.525) / (0.0821)(363.15)

n = 0.0175 moles of air

moles of oxygen = 20.95% x 0.0175 moles

moles of oxygen = 0.00367 moles

C) combustion of octane (C₈H₁₈) is as follows:

2C₈H₁₈ + 25O₂ ==> 16CO₂ + 18H₂O .. balanced equation for combustion of octane

molar mass C₈H₁₈ = 114.2 g / mole

grams C₈H₁₈ burned in part (B) = 0.00367 mol O₂ x 2 mol C₈H₁₈ / 25 mol O₂ x 114 g / mol =

0.0335 g C₈H₁₈