Answer:
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
