Question

Isooctane, C8H18, is the component of gasoline from which the term octane rating derives.

Answer 1

Answer:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) Mass of CO₂ produced annually from this combustion of isooctane gasoline = 1.12 × 10⁵ Kg

c) CO₂ produced from the combustion of the gasoline in a year will occupy 5.632 × 10⁷ L

d) There needs to be a minimum of 1.752 × 10⁷ moles of air and 3.92 × 10⁸ L of air for the oxygen to be in excess all through the year of gasoline combustion.

Explanation:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) C₈H₁₈ has a density of 0.792 mg/L.

Since density = mass/volume;

mass = density × volume

Mass of C₈H₁₈ with 4.6 x 10^10 L volume = 0.792 × 4.6 x 10^10 = 3.643 × 10^10 mg = 3.643 × 10⁷ g.

To obtain the mass of CO₂ produced, we need the number of moles of C₈H₁₈ that burned.

Number of moles = mass/molar mass

Molar mass of C₈H₁₈ = (8×12) + 18 = 114g/mol

Number of moles of C₈H₁₈ = (3.643 × 10⁷)/114 = (3.2 × 10⁵) moles.

From the chemical reaction,

1 mole of C₈H₁₈ burns to give 8 moles of CO₂

(3.2 × 10⁵) moles will give 8 × 3.2 × 10⁵ = (2.56 × 10⁶) moles of CO₂

Mass of CO₂ produced = number of moles × Molar mass

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ produced = 2.56 × 10⁶ × 44 = 1.12 × 10⁸ g = 1.12 × 10⁵ kg

c) 1 mole of any gas at stp occupies 22.4L

2.56 × 10⁶ moles of CO₂ will occupy 2.56 × 10⁶ × 22.4 = 5.632 × 10⁷ L

d) 1 mole of C₈H₁₈ requires 23/2 moles of O₂ for complete combustion yearly.

3.2 × 10⁵ moles would require 3.2 × 10⁵ × 23/2 = 3.68 × 10⁶ moles of O₂

O₂ makes up 21% of the air

That is,

0.21 moles of O₂ would be contained in 1 mole of air

3.68 × 10⁶ moles of O₂ would be contained in (3.68 × 10⁶ × 1)/0.21 = 1.752 × 10⁷ moles of air.

1 mole of any gas at stp occupies 22.4L

1.752 × 10⁷ of air will occupy

1.752 × 10⁷ × 22.4/1 = 3.92 × 10⁸ L of air!