Problem 85 - Chemistry Matter and Change

Question

Grilling what mass of propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) must be burned in a barbecue grill to release 4560 \(\mathrm{kJ}\) of heat? The \(\Delta H_{\mathrm{comb}}\) of propane is \(-2219 \mathrm{kJ} / \mathrm{mol}\)

Step-by-Step Solution

Answer

To release 4560 kJ, approximately 90.5 g of propane needs to be burned.

Step 1: Understand the Problem

We need to find out how much mass of propane (\(\mathrm{C}_3\mathrm{H}_8\)) is burned to release 4560 kJ of heat. Given the heat of combustion for propane is \(\Delta H_{\mathrm{comb}} = -2219 \mathrm{kJ/mol}\).

Step 2: Calculate Moles of Propane Combusted

The heat of combustion (\(\Delta H_{\mathrm{comb}}\)) tells us how much energy is released per mole of propane combusted. We can use the formula:\[n = \frac{\text{desired energy release}}{\Delta H_{\mathrm{comb}}}\]Plugging in the values: \[n = \frac{4560 \ \mathrm{kJ}}{-2219 \ \mathrm{kJ/mol}}\]Calculate \(n\).\[n \approx -2.054 \ \mathrm{mol}\]

Step 3: Calculate Mass of Propane Required

To find the mass, use the formula:\[\text{mass} = n \times \text{molar mass of propane}\]The molar mass of propane (\(\mathrm{C}_3\mathrm{H}_8\)) is calculated as:\[3 \times 12.01 \ \mathrm{g/mol} + 8 \times 1.01 \ \mathrm{g/mol} = 44.09 \ \mathrm{g/mol}\]Thus:\[\text{mass} = 2.054 \ \mathrm{mol} \times 44.09 \ \mathrm{g/mol}\]Calculate the mass.\[\text{mass} \approx 90.5 \ \mathrm{g}\]

Key Concepts

Propane Combustion

When it comes to grilling, understanding the combustion of propane is crucial. Propane combustion refers to the chemical reaction where propane (\(\text{C}_3\text{H}_8\)) reacts with oxygen to produce carbon dioxide, water, and energy. This process is vital to fuel many everyday activities like cooking and heating.

In the reaction, propane is represented by the chemical formula \(\text{C}_3\text{H}_8\). When burned, it combines with oxygen (\(\text{O}_2\)) in the air. This process releases a significant amount of energy, which is why it's so useful for grilling:\[\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} + \text{Energy}\]For every mole of propane burned, a standard amount of energy is released, known as the heat of combustion. For propane, this is a specific value of \(-2219 \text{kJ/mol}\), meaning this amount of energy is released along with the products. The negative sign indicates that the process is exothermic, releasing energy to the surroundings.

Knowing this heat value is essential to calculate how much propane is needed to produce a desired amount of energy.

Energy Calculation

Energy calculation in propane combustion helps us determine how much propane is needed to generate a specific amount of energy. This is essential, especially when trying to determine resource requirements for applications such as grilling or heating a space.

To calculate the moles of propane needed, employ the energy release formula, which relates energy to the heat of combustion:

• Formula: \[n = \frac{\text{desired energy release}}{\Delta H_{\text{comb}}}\]
• Where \(n\) is the number of moles of propane, \(\Delta H_{\text{comb}}\) is the heat of combustion, and the desired energy release is the needed energy in kilojoules.

In the exercise, we have 4560 kJ as the desired energy release and \(-2219 \text{kJ/mol}\) as the heat of combustion. When calculating using the formula:\[n = \frac{4560 \ \text{kJ}}{-2219 \ \text{kJ/mol}}\approx -2.054\text{ mol}\]We see that a little over 2 moles of propane are necessary to release the required energy. Although the mole calculation results in a negative due to \(\Delta H_{\text{comb}}\), it just backs the direction of energy flow as releasing.

Molar Mass Calculation

To find out how much propane is needed in terms of mass, it's important to understand how to calculate the molar mass. The molar mass of a compound tells us the mass of one mole of that compound, which is quite useful in stoichiometry and energy calculations.

For propane (\(\text{C}_3\text{H}_8\)), calculating the molar mass involves summing the atomic masses of all atoms in the molecule.

• Carbon (\(\text{C}\)) contributes \(12.01 \text{ g/mol}\) per atom and there are 3 carbon atoms in propane.
• Hydrogen (\(\text{H}\)) contributes \(1.01 \text{ g/mol}\) per atom, with 8 hydrogen atoms present.

Thus, the calculation is:\[\text{Molar Mass of } \text{C}_3\text{H}_8 = (3 \times 12.01 \text{ g/mol}) + (8 \times 1.01 \text{ g/mol}) = 44.09 \text{ g/mol}\]Once we have our moles calculated from the energy release, we use this molar mass to find the actual mass:\[\text{mass} = 2.054 \ \text{mol} \times 44.09 \ \text{g/mol} = 90.5 \ \text{g}\]This tells us that approximately 90.5 grams of propane are needed to release 4560 kJ of energy, showing us how molar mass connects to practical uses.