how many grams of propane do you need to combust 525 kJ of heat
To determine the number of grams of propane needed to combust 525 kJ of heat, we need to use the concept of energy conversion and the molar mass of propane.
Here's how you can calculate it step by step:
1. Determine the molar mass of propane (C3H8):
- Propane consists of three carbon atoms (C) and eight hydrogen atoms (H).
- The molar mass of carbon (C) is approximately 12.01 g/mol.
- The molar mass of hydrogen (H) is approximately 1.01 g/mol.
- Multiply the molar mass of carbon by 3 (number of carbon atoms) and add it to the molar mass of hydrogen multiplied by 8 (number of hydrogen atoms) to get the molar mass of propane:
Molar mass of propane (C3H8) = (3 * 12.01 g/mol) + (8 * 1.01 g/mol) = 44.11 g/mol
2. Convert the given energy from kilojoules (kJ) to joules (J):
- 1 kJ = 1000 J
- Therefore, 525 kJ = 525,000 J
3. Convert the given energy to moles of propane using the molar enthalpy of combustion:
- The molar enthalpy of combustion of propane is -2220 kJ/mol. This means that 2220 kJ of energy is released when one mole of propane is combusted.
- Divide the energy in joules (525,000 J) by the molar enthalpy of combustion (-2220 kJ/mol) to get the number of moles of propane:
Moles of propane = (525,000 J) / (-2220 kJ/mol) = -0.236 moles
4. Convert moles of propane to grams using the molar mass of propane:
- Multiply the number of moles of propane by the molar mass of propane:
Grams of propane = (-0.236 moles) * (44.11 g/mol) = -10.37 g
Note: The negative sign indicates that the combustion of 525 kJ of heat requires 10.37 grams of propane.
Therefore, you would need approximately 10.37 grams of propane to combust 525 kJ of heat.