Question

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

A. Write a balanced equation for the combustion of isooctane to yield CO2 and H2O.

B. Assuming that gasoline is 100% isooctane, the isooctane burns to produce only CO2 and H2O, and that the density of isooctane is 0.792 g/mL, what mass of CO2 in kilograms is produced each year by the annual US gasoline consumption of 4.6 x 10^10 L?

C. What is the volume in L of this CO2 at STP?

D. How many moles of air are necessary for the combustion of 1 mol of isooctane, assuming that air is 21.0% O2 by volume.? What is volume in L of this air at STP?

Answer 1

The balanced equation for the combustion of isooctane is C8H18(l) + 11O2(g) → 8CO2(g) + 9H2O(g). Assuming that gasoline is 100% isooctane, the mass of CO2 produced each year by the annual US gasoline consumption of 4.6 x 10^10 L is 9.04 x 10^6 kg. The volume of this CO2 at STP is 2.03 x 10^11 L. We use the molar ratios from the balanced equation to determine the number of moles of air necessary for the combustion of 1 mol of isooctane. The volume in L of this air at STP is 1173 L.

Step-by-step explanation:

The balanced chemical equation for the combustion of isooctane (C8H18) is:

C8H18(l) + 11O2(g) → 8CO2(g) + 9H2O(g)

Assuming that gasoline is 100% isooctane, we can use the balanced equation to calculate the mass of CO2 produced each year by the annual US gasoline consumption of 4.6 x 1010 L. We start by calculating the mass of isooctane consumed, then use the molar ratios from the balanced equation to calculate the mass of CO2 produced. Given that the density of isooctane is 0.792 g/mL, the mass consumed is 4.6 x 1010 L x 0.792 g/mL = 3.64 x 1010 g. The molar mass of CO2 is 44.01 g/mol, so the number of moles of CO2 produced is 28 x 3.64 x 1010 g / 114.22 g/mol = 9.04 x 109 mol. Finally, we convert to kilograms by dividing by 1000, so the CO2 produced is 9.04 x 106 kg.

To calculate the volume of CO2 at STP, we use the ideal gas law. At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. So, the volume of CO2 at STP is 9.04 x 109 mol x 22.4 L/mol = 2.03 x 1011 L.

We use the molar ratios from the balanced equation to determine the number of moles of air necessary for the combustion of 1 mol of isooctane. From the equation, we can see that 11 moles of O2 are required for the combustion of 1 mol of isooctane. Since air is 21.0% O2 by volume, the volume of air required is 11 / 0.21 = 52.4 mol. At STP, 1 mole of gas occupies 22.4 L, so the volume of air required is 52.4 mol x 22.4 L/mol = 1173 L.

Answer 2

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.

Step-by-step 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!