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)

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:

Given that:

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

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:

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

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:

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Explanation:

1. Number of moles of gasoline

a)  Convert 60.0 liters to grams

b) Convert 46,200 grams to moles

2. Number of moles of carbon dioxide, CO₂ produced

a) Balanced chemical equation (given):

b) mole ratio:

Solve for x:

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

Use the ideal gas equation:

Substitute and compute:

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

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