Question

The density of pentane, a component of gasoline, is \(0.63 \mathrm{~g} / \mathrm{mL}\). The heat of combustion for pentane is \(845 \mathrm{kcal} / \mathrm{mole}\). a. Write the equation for the complete combustion of pentane. b. What is the molar mass of pentane? c. How much heat is produced when 1 gallon of pentane is burned \((1 \mathrm{gal}=4 \mathrm{qt}) ?\) d. How many liters of \(\mathrm{CO}_{2}\) at STP are produced from the complete combustion of 1 gallon of pentane?

Short answer

The heat produced is approximately 27902.35 kcal. The volume of CO₂ produced is approximately 3707.36 L.

Step-by-step solution

Write the combustion equation

For complete combustion of pentane (\text{C}_5\text{H}_{12}), the equation is: \[\text{C}_5\text{H}_{12} + 8\text{O}_2 \rightarrow 5\text{CO}_2 + 6\text{H}_2\text{O}\]

Calculate the molar mass of pentane

Calculate the molar mass by summing the atomic masses of its constituent atoms. Pentane (\text{C}_5\text{H}_{12}) has 5 carbon atoms and 12 hydrogen atoms: \[ \text{M}(\text{C}_5\text{H}_{12}) = 5(12.01 \text{ g/mol}) + 12(1.01 \text{ g/mol}) = 60.05 \text{ g/mol} + 12.12 \text{ g/mol} = 72.17 \text{ g/mol}\]

Determine the volume of 1 gallon in mL

Convert 1 gallon to milliliters (mL). Since 1 gallon = 4 quarts and 1 quart = 946 mL: \[ 1 \text{ gal} \times 4 \text{ qt/gal} \times 946 \text{ mL/qt} = 3784 \text{ mL}\]

Calculate the mass of 1 gallon of pentane

Using the density (\text{density} = 0.63 \text{ g/mL}) and the volume calculated: \[ \text{mass} = \text{volume} \times \text{density} = 3784 \text{ mL} \times 0.63 \text{ g/mL} = 2383.92 \text{ g}\]

Convert the mass of pentane to moles

Using the molar mass calculated: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{2383.92 \text{ g}}{72.17 \text{ g/mol}} \approx 33.03 \text{ moles}\]

Calculate the heat produced from burning 1 gallon of pentane

Using the heat of combustion value (\text{combustion} = 845 \text{ kcal/mole}): \[ \text{heat produced} = 33.03 \text{ moles} \times 845 \text{ kcal/mole} \approx 27902.35 \text{ kcal}\]

Calculate the volume of CO₂ produced at STP

From the combustion equation, 1 mole of pentane produces 5 moles of CO₂. Thus, 33.03 moles of pentane produce: \[ 33.03 \text{ moles} \times 5 = 165.15 \text{ moles} \text{ of CO}_2\] At STP, 1 mole of a gas occupies 22.4 liters. So, the volume of CO₂ gas produced is: \[ 165.15 \text{ moles} \times 22.4 \text{ L/mole} \approx 3707.36 \text{ L}\]

Key concepts

Pentane Combustion Equation

The equation for the combustion of pentane reflects how pentane reacts with oxygen to produce carbon dioxide and water. In basic terms, pentane (a hydrocarbon with the formula \text{C}_5\text{H}_{12}) combines with oxygen (\text{O}_2) from the air. This reaction releases energy in the form of heat and produces carbon dioxide (\text{CO}_2) and water (\text{H}_2\text{O}). The balanced equation for this reaction is:

\[\text{C}_5\text{H}_{12} + 8\text{O}_2 \rightarrow 5\text{CO}_2 + 6\text{H}_2\text{O}\]

This reaction is an example of a complete combustion, meaning all of the pentane reacts with oxygen.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of that substance measured in grams per mole (g/mol). To calculate the molar mass of pentane (\text{C}_5\text{H}_{12}), you sum the atomic masses of all the atoms in the molecule. Pentane is made up of 5 carbon atoms and 12 hydrogen atoms. The atomic mass of carbon is 12.01 g/mol and that of hydrogen is 1.01 g/mol.

So, the molar mass calculation of pentane is:

\[ 5(12.01 \text{ g/mol}) + 12(1.01 \text{ g/mol}) = 60.05 \text{ g/mol} + 12.12 \text{ g/mol} = 72.17 \text{ g/mol} \]

Density and Volume Conversion

To find out how much a liquid weighs, knowing its density and volume is key. Density tells you how much mass is in a specific volume, usually in grams per milliliter (g/mL). For pentane, its density is 0.63 g/mL. If you need to find out how much 1 gallon of pentane weighs, you must first convert gallons to milliliters.

1 gallon = 4 quarts, and 1 quart = 946 mL. Thus, 1 gallon is:

\[ 1 \text{ gal} \times 4 \text{ qt/gal} \times 946 \text{ mL/qt} = 3784 \text{ mL} \]

Now, you can find the mass by multiplying the volume by the density:

\[ 3784 \text{ mL} \times 0.63 \text{ g/mL} = 2383.92 \text{ g} \]

Heat of Combustion

The heat of combustion is the amount of heat energy released when a substance completely burns in the presence of oxygen. For pentane, the heat of combustion is 845 kcal per mole. To find out how much heat is produced when 1 gallon of pentane is burned, you first convert the mass of pentane into moles using its molar mass.

Using the mass (2383.92 g) and the molar mass (72.17 g/mol):

\[ \text{moles} = \frac{2383.92 \text{ g}}{72.17 \text{ g/mol}} \approx 33.03 \text{ moles} \]

Then, you multiply the number of moles by the heat of combustion:

\[ 33.03 \text{ moles} \times 845 \text{ kcal/mole} \approx 27902.35 \text{ kcal} \]

So, 1 gallon of pentane produces about 27902.35 kilocalories of heat.

STP Gas Volume Calculation

Standard Temperature and Pressure (STP) refer to a set of conditions for measuring gases: a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (atm). At STP, 1 mole of any gas occupies 22.4 liters. To find out how much carbon dioxide (\text{CO}_2) is produced when 1 gallon of pentane combusts, you use the balanced combustion equation.

From the equation, 1 mole of pentane produces 5 moles of carbon dioxide:

\[ 33.03 \text{ moles of pentane} \times 5 = 165.15 \text{ moles of CO}_2 \]

At STP, this amount of \(\text{CO}_2\) translates into volume as follows:

\[ 165.15 \text{ moles} \times 22.4 \text{ L/mole} \approx 3707.36 \text{ L} \]

Hence, the combustion of 1 gallon of pentane produces approximately 3707.36 liters of carbon dioxide at STP.