How many grams of propane (C3H8) must be burned to release 3600kJ of energy?

Standard Heat of combustion for Propane=
-2220 kJ/Mole

To determine the number of grams of propane that must be burned to release 3600 kJ of energy, we need to use the molar mass of propane (C3H8) and the given standard heat of combustion.

First, let's calculate the number of moles of propane required:

1. Determine the molar mass of propane:
C (carbon) has a molar mass of 12.01 g/mol, and there are 3 carbons in propane, so the carbon part has a molar mass of 12.01 g/mol x 3 = 36.03 g/mol.
H (hydrogen) has a molar mass of 1.01 g/mol, and there are 8 hydrogens in propane, so the hydrogen part has a molar mass of 1.01 g/mol x 8 = 8.08 g/mol.
Adding both parts together: 36.03 g/mol + 8.08 g/mol = 44.11 g/mol.

2. Calculate the number of moles of propane:
We can use the equation: moles = energy (kJ) / heat of combustion (kJ/mol).
Thus, moles = 3600 kJ / (-2220 kJ/mol).
Simplifying: moles = -1.62 mol.

Since moles can't be negative, it indicates that the heat provided is negative (released). So, to release 3600 kJ of energy, approximately 1.62 moles of propane need to be burned.

Finally, to convert moles to grams, we multiply by the molar mass of propane:
1.62 mol x 44.11 g/mol = approximately 71.40 g

Therefore, approximately 71.40 grams of propane must be burned to release 3600 kJ of energy.