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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?

Respuesta :

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Answer:

A. 2C₈H₁₈ + 25O₂  →  16CO₂  +  18H₂O

B. 1.12×10¹¹ kg

C. 5.72×10¹² liters

Explanation:

The reaction for the combustion is:

2C₈H₁₈ + 25O₂  →  16CO₂  +  18H₂O

Density of gasoline help us to determine the mass used, and afterwards the moles (mass / molar mass).

Gasoline density = Gasoline mass / Gasoline volume.

We must convert the volume in L → 4.6×10¹⁰ L . 1 mL / 1×10⁻³ L = 4.6×10¹³ mL

0.792 g/mL = Gasoline mass / 4.6×10¹³ mL

0.792 g/mL . 4.6×10¹³ mL = Gasoline mass → 3.64×10¹³ g

Moles of gasoline → 3.64×10¹³ g / 114 g/mol = 3.19×10¹¹ moles

Let's determine the CO₂ produced. Ratio is 2:16

2 moles of Isooctane produce 16 moles of CO₂

Therefore, 3.19×10¹¹ moles would produce (3.19×10¹¹ moles  .16)/2 = 2.55×10¹² moles of CO₂

If we want to know the mass, let's convert it. (mol . molar mass)

2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g

Let's convert the g to kg → 1.12×10¹⁴ g . 1kg / 1000 g = 1.12×10¹¹ kg

Let's calculate the volume with the Ideal Gases Law

P . V = n . R . T

At STP, pressure is 1atm and T is 273K

1 atm . V = 2.55×10¹² mol . 0.082 L.atm/mol.K . 273K

V = (2.55×10¹² mol . 0.082 L.atm/mol.K . 273K) / 1 atm → 5.72×10¹² liters