Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is \(\mathrm{C}_{12} \mathrm{H}_{26}\) (a) Write a balanced equation, using the simplest whole-number coefficients, for the complete combustion of kerosene to gases. (b) If \(\Delta H_{\mathrm{rn}}^{\circ}=-1.50 \times 10^{4} \mathrm{~kJ}\) for the combustion equation as written in part (a), determine \(\Delta H_{\mathrm{f}}^{\circ}\) of kerosene. (c) Calculate the heat released by combustion of 0.50 gal of kerosene \((d\) of kerosene \(=0.749 \mathrm{~g} / \mathrm{mL})\) (d) How many gallons of kerosene must be burned for a kerosene furnace to produce \(1250 .\) Btu \((1 \mathrm{Btu}=1.055 \mathrm{~kJ}) ?\)

The first step in solving any combustion problem is to write the balanced chemical equation. This involves hydrocarbons reacting with oxygen to produce carbon dioxide and water. For this exercise, kerosene's average chemical formula is given as \(\text{C}_{12}\text{H}_{26}\). The unbalanced equation is:
\[\text{C}_{12}\text{H}_{26} + O_2 \rightarrow CO_2 + H_2O\]
To balance it, we start by balancing the carbons, then the hydrogens, and finally the oxygens. The balanced equation is:
\[\text{C}_{12}\text{H}_{26} + 18.5O_2 \rightarrow 12 CO_2 + 13 H_2O\]
This balanced equation ensures that for every molecule of kerosene, 18.5 molecules of oxygen are required, yielding 12 molecules of carbon dioxide and 13 molecules of water.

Understanding the enthalpy of formation (\( \Delta H_{f}^{o}\)) is crucial for calculating the heat produced or consumed in chemical reactions. The enthalpy of formation of a compound is the heat change when one mole of a compound is formed from its elements in their standard states.
In this exercise, you are given the reaction enthalpy for the combustion of kerosene (\( \Delta H_{rn}^{o} =-1.50 \times 10^{4} \text{kJ}\)). The goal is to find the enthalpy of formation for kerosene. The combustion reaction is:
\[ \text{C}_{12} \text{H}_{26} +18.5 \text{O}_{2} \rightarrow 12 \text{CO}_{2} + 13 \text{H}_{2} \text{O} \]
Using the enthalpy of formation for \( \text{CO}_{2} \) (-393.5 kJ/mol) and \( \text{H}_{2} \text{O} \) (-241.8 kJ/mol), we can apply Hess’s Law to find the unknown \( \Delta H_{f}^{o} \text{ of kerosene} \).
Thus,
\[ \Delta H_{f}^{o} (\text{C}_{12}\text{H}_{26}) = \frac{1.50 \times 10^{4} \text{kJ} - [12(-393.5) + 13(-241.8)]}{1} \] This allows us to solve for the unknown enthalpy.

To calculate the heat released by burning a specific amount of kerosene, we need to follow unit conversion steps and use density and molecular weight data. First, convert the volume of kerosene from gallons to milliliters (1 gal = 3,785.41 mL).
For 0.50 gal: \[ 0.50 \text{ gal} \times 3785.41 \text{ mL/gal} = 1892.705 \text{ mL} \]
Use the density of kerosene (0.749 g/mL) to find its mass: \[ 1892.705 \text{ mL} \times 0.749 \text{ g/mL} = 1417.17 \text{ g} \]
Next, convert the mass to moles using the molar mass of \( \text{C}_{12}\text{H}_{26} \) (170 g/mol): \[ \text{moles of kerosene} = \frac{1417.17 \text{ g}}{170 \text{ g/mol}} = 8.34 \text{ moles} \]
Finally, calculate the heat released for the combustion: \[ 8.34 \text{ moles} \times -1.50 \times 10^{4} \text{ kJ/mol} = -1.25 \times 10^{3} \text{ kJ} \]

Unit conversions are essential in chemistry for comparing different measurements. Let's start with converting energy units from Btu to kJ (1 Btu = 1.055 kJ). For a kerosene furnace to produce 1250 Btu: \[ 1250 \text{ Btu} \times 1.055 \text{ kJ/Btu} = 1318.75 \text{ kJ} \]
Use the heat released per mole of kerosene to determine the moles needed: \[ \text{moles of kerosene} = \frac{1318.75 \text{ kJ}}{1.50 \times 10^{4} \text{ kJ/mol}} = 0.088 \text{ moles} \]
Convert moles to mass: \[ \text{mass of kerosene} = 0.088 \text{ moles} \times 170 \text{ g/mol} = 15 \text{ g} \]
Convert the mass to volume using the density: \[ \text{Volume (mL)} = \frac{15 \text{ g}}{0.749 \text{ g/mL} } = 20 \text{ mL} \]
Finally, convert milliliters to gallons: \[ \text{Volume (gal)} = \frac{20 \text{ mL}}{3785.41 \text{ mL/gal} } = 0.005 \text{ gal} \]
These conversions ensure accurate and applicable results for practical applications.