Questions: The combustion enthalpy of cyclobutane, C4 H8, is 2.721 x 10^3 kJ / mol. Write a balanced equation for the complete combustion of cyclobutane. Calculate how much energy is released during the complete combustion of 414 g cyclobutane.

Solution Steps

The complete combustion of a hydrocarbon involves reacting it with oxygen (\(\text{O}_2\)) to produce carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). The general form for the combustion of cyclobutane (\(\text{C}_4\text{H}_8\)) is:

\[ \text{C}_4\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

To balance the equation, we need to ensure that the number of each type of atom is the same on both sides of the equation:

• Carbon: 4 atoms in \(\text{C}_4\text{H}_8\) require 4 \(\text{CO}_2\).
• Hydrogen: 8 atoms in \(\text{C}_4\text{H}_8\) require 4 \(\text{H}_2\text{O}\).
• Oxygen: 4 \(\text{CO}_2\) and 4 \(\text{H}_2\text{O}\) require 12 \(\text{O}\) atoms, which means 6 \(\text{O}_2\).

The balanced equation is:

\[ \text{C}_4\text{H}_8 + 6\text{O}_2 \rightarrow 4\text{CO}_2 + 4\text{H}_2\text{O} \]

First, calculate the molar mass of cyclobutane (\(\text{C}_4\text{H}_8\)):

• Carbon: \(4 \times 12.01 \, \text{g/mol} = 48.04 \, \text{g/mol}\)
• Hydrogen: \(8 \times 1.008 \, \text{g/mol} = 8.064 \, \text{g/mol}\)

Total molar mass of \(\text{C}_4\text{H}_8\) is \(48.04 + 8.064 = 56.104 \, \text{g/mol}\).

Now, calculate the moles of cyclobutane in 414 g:

\[ \text{Moles of } \text{C}_4\text{H}_8 = \frac{414 \, \text{g}}{56.104 \, \text{g/mol}} \approx 7.379 \, \text{mol} \]

The combustion enthalpy of cyclobutane is given as \(2.721 \times 10^3 \, \text{kJ/mol}\). Therefore, the energy released during the combustion of 7.379 moles of cyclobutane is:

\[ \text{Energy released} = 7.379 \, \text{mol} \times 2.721 \times 10^3 \, \text{kJ/mol} \approx 20,080 \, \text{kJ} \]

Final Answer