Question

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)

Answer 1

Answer:

• 79,000 liters

Explanation:

1. Number of moles of gasoline

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

2. Number of moles of carbon dioxide, CO₂ produced

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

3. Convert the number of moles of carbon dioxide to volume

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