Question

What volume of air is needed to burn an entire 55-L (approximately 15-gal) tank of gasoline? Assume that the gasoline is pure octane, C3H18. Hint: Air is 20% oxygen, 1 mol of a gas occupies about 25 L at room temperature, and the density of octane is 0.70 g/cm

Answer 1

Answer:

The volume of air required is 527,686.25L.

Explanation:

When the question says "burn", it refers to a combustion reaction, where a substance (in this case octane) reacts with oxygen to produce carbon dioxide and water.

Step 1: Write a balanced equation

Considering it is a combustion reaction, the balanced equation is:

C₈H₁₈ + 12.5 O₂ ⇄ 8 CO₂ + 9 H₂O

In this step, we start balancing elements that are present only in one compound on each side of the equation, namely, carbon and hydrogen.

To finish, it is important to count that there are the same number of atoms on both sides of the equation. In this case there are 8 atoms of Carbon, 18 atoms of Hydrogen and 25 atoms of Oxygen, so it is balanced.

Step 2: Find out the mass of C₈H₁₈

Since the balanced equation gives us information about the mass of C₈H₁₈ involved in the reaction, we need to find out how many grams we have.

The info we have is:

• 55 L of gasoline (assuming gasoline to be pure octane).
• The density of octane is 0,70 g/cm³

Density relates mass and volume, so we can find out how many grams are represented by 55 L. Since the units used are different, first we need to convert liters into cm³. We use the conversión factor 1 L = 1000 cm³.

$55L.\frac{1000cm^{3} }{1L} =55000cm^{3}$

Since density = mass/volume, we can solve for mass:

$mass = density.volume=0.70\frac{g}{cm^{3} } .55,000cm^{3} =38,500g$

Step 3: Establish the theoretical relationship between the mass of octane and the moles of oxygen.

This relationship comes from the balanced equation.

  For octane:

  molar mass of C₈H₁₈ = molar mass of C . 8 + molar mass of H . 18 =

  12 g/mol . 8 +  1g/mol . 18 = 114 g/mol

  According to the balanced equation reacts 1 mol of octane, which means      114 grams of it.

  For oxygen:

  According to the balanced equation, 12.5 moles of oxygen react.

Then, the relationship is 114 g octane : 12.5 moles of oxygen

Step 4: Use the theorethical relationship to find the moles of oxygen that reacted

We use the mass of octane found in step 2 and apply the proper conversión factor.

$38,500g (octane) . \frac{12.5mol (oxygen)}{114g (octane)} = 4,221.49mol(oxygen)$

Step 5: Find out the volume of oxygen.

We know that 1 mol of any gas at room temperature occupies about 25 L. Then,

$4,221.49mol(oxygen).\frac{25L(oxygen)}{1mol(oxygen)} =105,537.25 L (oxygen)$

Step 6: Look the volume of air that contains such amount of oxygen

Given oxygen represents 20% of air, we can use that relationship to find the volume of air.

$105,537.25L(oxygen).\frac{100L(air)}{20L(oxygen)} =527,686.25L (air)$