An automobile engine has a cylinder with a volume of 500.0 L that is filled with air (21.00 % oxygen) at a temperature of 55.00 C and a pressure of 2.000 atm. What is the mass of octane, C8H18 that must be injected to reacted with all of the oxygen in the cylinder to produce carbon dioxide and water?

The answer to the question is

35.64 grams of octane, C₈H₁₈ must be injected to reacted with all of the oxygen in the cylinder to produce carbon dioxide and water

Explanation:

Firstly we wrte the equation for the reaction as follows

2C₈H₁₈+25O₂ → 16CO₂+18H₂O

From where it is seen that 2 moles of octane combine with 25 moles of O₂ to form 16 moles of CO₂ and 18 moles of H₂O

However the conyainer contains

500.0 L of air containing 21.00 % oxygen

Which is 0.21 × 500 =  105 L = ‪0.105‬ m³ of oxygen gas

drom the combined  gas equation  we have

PV = nRT hence n = PV/(RT) where P = 2.000 atm = ‪202650‬ Pa, T = 55.00 C   = ‪328.15‬ K and R = 8.314 has units of J / (mol K).

21278.25/2728.24 = 7.8 moles of O₂

Therefore since 25 moles of O₂ react with 2 moles of C₈H₁₈ then

1 mole of O₂ will react with 1/25 moles of C₈H₁₈ this gives and 7.8 moles will react with 7.8/25 moles of C₈H₁₈

One mole of C₈H₁₈  weighs 114.23 g

Then 7.8/25 moles C₈H₁₈ weigh 35.64 grams

The mass of octane, C₈H₁₈ that must be injected to reacted with all of the oxygen in the cylinder to produce carbon dioxide and water is 35.64 grams

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.