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saveliy_v [14]
3 years ago
10

How many L of carbon dioxide at 1.00 atm and 298.15 K are released from a car's engine upon consumption of a 60.0 L LIQUID tank

gasoline? (Gasoline density: 0.77 kg/L, Molar mass of C₈H₁₈: 114.2 g/mol) C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)
Chemistry
1 answer:
torisob [31]3 years ago
3 0

Answer:

  • <u>79,000 liters</u>

Explanation:

<u>1. Number of moles of gasoline</u>

a)  Convert 60.0 liters to grams

  • density = 0.77kg/liter
  • density = mass / volume
  • mass = density × volume
  • mass = 0.77kg/liter × 60.0 liter = 46.2 kg

  • 46.2kg × 1,000g/kg = 46,200g

b) Convert 46,200 grams to moles

  • molar mass of C₈H₁₈ = 114.2 g/mol
  • number of moles = mass in grams / molar mass
  • number of moles = 46,200g / (114.2 gmol) = 404.55 mol

<u>2. Number of moles of carbon dioxide, CO₂ produced</u>

a) Balanced chemical equation (given):

  • C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)

b) mole ratio:

  • 1 mol C₈H₁₈ / 8 mol CO₂ = 404.55 mol C₈H₁₈ / x

Solve for x:

  • x = 404.55mol C₈H₁₈ × 8 mol CO₂ / 1mol C₈H₁₈ = 3,236.4 mol CO₂

<u> 3. Convert the number of moles of carbon dioxide to volume</u>

Use the ideal gas equation:

  • pV = nRT
  • V = nRT/p
  • p = 1 atm
  • T = 298.15K
  • n = 3,236.4 mol
  • R = 0.08206 (mol . liter)/ (K . mol)

Substitute and compute:

  • V =3,236.4 mol × 0.08206 (mol . liter) / (K . mol) 298.15K / 1 atm
  • V = 79,183 liter

Round to two significant figures (because the density has two significant figures): 79,000 liters ← answer

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