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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)

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Edufirst

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

ajeigbeibraheem

The liters L of carbon dioxide at 1.00 atm and 298.15 K released from the car's engine upon consumption of a 60.0 L LIQUID tank gasoline is 79230.76 L

From the given information;

The equation for the reaction can be expressed as:

[tex]\mathbf{C_8H_8 _{(l)} + \frac{25}{2}O_2_{(g)} \to 8 CO_2_{(g)} + 9H_2O_{(g)}}[/tex]

Given that:

  • Pressure = 1.00 atm
  • Temperature = 298.15 K
  • amount of gasoline used = 60.0 L
  • gasoline density = 0.77 kg/L
  • Molar mass of C₈H₁₈: 114.2 g/mol

The first thing to do is to determine the mass amount of gasoline used by using the relation:

[tex]\mathbf{Density = \dfrac{mass}{volume}}[/tex]

[tex]\mathbf{0.77 \ kg/L = \dfrac{mass}{60 \ L}}}[/tex]

mass amount of gasoline used = 0.77 kg/L × 60 L

mass amount of gasoline used = 46.2 kg

mass amount of gasoline used = 46.2 × 1000g

mass amount of gasoline used = 46200 g

Now, since we know the mass amount of the gasoline used, we can determine the number of moles of the gasoline by using the formula:

[tex]\mathbf{number of moles = \dfrac{mass}{molar \ mass}}[/tex]

[tex]\mathbf{number of moles = \dfrac{46200 \ g}{114.2 \ g/mol}}[/tex]

[tex]\mathbf{number \ of \ moles = 404.6 \ moles }[/tex]

From the given reaction, 1 mole of gasoline react to produce 8 moles of CO₂

As such, 404.6 moles of gasoline will produce = (404.6 moles × 8)/ 1

= 3236.8 moles of CO₂ is produced.

Now, using the ideal gas equation to determine the volume of CO₂ that  are released;

PV = nRT

[tex]\mathbf{1.00 atm \times V = 3236.8 moles \times 0.0821 atm L /K/mol \times 298.15 \ K}[/tex]

[tex]\mathbf{V = \dfrac{3236.8 moles \times 0.0821 atm L /K/mol \times 298.15 \ K}{1.00 atm}}[/tex]

V = 79230.76 L

Therefore, we can conclude that the liters L of carbon dioxide at 1.00 atm and 298.15 K released from the car's engine upon consumption of a 60.0 L LIQUID tank gasoline is 79230.76 L

Learn more about ideal gas equation here:

https://brainly.com/question/18518493?referrer=searchResults

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