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Assuming that gasoline is 100% isooctane, that 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 U. S. gasoline consumption of 4.6×1010L?

Answer :

dsdrajlin

Answer:

1.12 × 10¹¹ kg of CO₂

Explanation:

Let's consider the combustion of isooctane.

C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(l)

We can establish the following relations:

  • 1 mL of isooctane has a mass of 0.792 g
  • The molar mass of isooctane is 114 g/mol
  • 1 mol of isooctane produces 8 moles of CO₂
  • The molar mass of CO₂ is 44.0 g/mol

Then, for 4.6 × 10¹⁰ L of gasoline (assuming that gasoline is 100% isooctane):

[tex]4.6\times 10^{10} LC_{8}H_{18}.\frac{10^{3}mLC_{8}H_{18} }{1LC_{8}H_{18}} .\frac{0.792gC_{8}H_{18}}{1mLC_{8}H_{18}} .\frac{1molC_{8}H_{18}}{114gC_{8}H_{18}} .\frac{8molCO_{2}}{1molC_{8}H_{18}} .\frac{44.0 \times 10^{-3}kgCO_{2}  }{1molCO_{2}} =1.12\times 10^{11} kgCO_{2}[/tex]

Eduard22sly

The mass of CO₂ produced each year by the annual U.S gasoline consumption of 4.6×10¹⁰ L is 1.12×10¹¹ Kg

We'll begin by calculating the mass of isooctane, C₈H₁₈. This can be obtained as follow:

Volume of C₈H₁₈ = 4.6×10¹⁰ L = 4.6×10¹⁰ × 1000 = 4.6×10¹³ mL

Density = 0.792 g/mL

Mass of C₈H₁₈ =?

Mass = Density × Volume

Mass of C₈H₁₈ = 0.792 × 4.6×10¹³

Mass of C₈H₁₈ = 3.6432×10¹³ g

Divide by 1000 to express in Kg

Mass of C₈H₁₈ = 3.6432×10¹³ / 1000

Mass of C₈H₁₈ = 3.6432×10¹⁰ Kg

  • Next, we shall determine mass of C₈H₁₈ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

2C₈H₁₈(l) + 25O₂(g) —> 16CO₂(g) + 18H₂O(l)

Molar mass of C₈H₁₈ = (8×12) + (1×18) = 114 g/mol

Mass of C₈H₁₈ from the balanced equation = 2 × 114 = 228 g

Divide by 1000 to express in Kg

228 / 1000 = 0.228 Kg

Molar mass of CO₂ = 12 + (16×2) = 44 g/mol

Mass of CO₂ from the balanced equation = 16 × 44 = 704 g

Divide by 1000 to express in Kg

704 / 1000 = 0.704 Kg

SUMMARY

From the balanced equation above,

0.228 Kg of C₈H₁₈ reacted to produce 0.704 Kg of CO₂

  • Finally, we shall determine the mass of CO₂ produced from 3.6432×10¹⁰ Kg of C₈H₁₈. This can be obtained as follow:

From the balanced equation above,

0.228 Kg of C₈H₁₈ reacted to produce 0.704 Kg of CO₂

Therefore,

3.6432×10¹⁰ Kg of C₈H₁₈ will react to produce = [tex]\frac{3.6432*10^{10} * 0.704}{0.228} \\\\[/tex] = 1.12×10¹¹ Kg of CO₂

Thus, the mass of CO₂ produced each year by the annual U.S gasoline consumption of 4.6×10¹⁰ L is 1.12×10¹¹ Kg

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