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
