Final answer:
The heat transferred when burning 84 grams of propane is approximately -4225.28 kJ, calculated by first determining the moles of propane in 84 grams and then multiplying by the given enthalpy change per mole, ΔH.
Step-by-step explanation:
To calculate the heat transferred when 84 grams of propane (C₃H₈) is burned, we need to use the given ΔH value of -2220 kJ, which is the enthalpy change per mole of propane combusted. First, we calculate the number of moles of propane in 84 grams using the molar mass of propane (C₃H₈), which is 44.09 g/mol. Once we have the number of moles, we can multiply it by the enthalpy change per mole to find the total heat transferred.
The calculation is as follows:
- Numerator of moles of propane = mass (g) / molar mass (g/mol)
- Moles of propane = 84 g / 44.09 g/mol
- Heat transferred (Q) = moles of propane × ΔH
Performing these calculations:
- Moles of propane = 84 g / 44.09 g/mol = 1.904 moles (approx)
- Heat transferred (Q) = 1.904 moles × (-2220 kJ/mol) = -4225.28 kJ
Therefore, the heat transferred when burning 84 grams of propane is approximately -4225.28 kJ.