Answer:
The correct answer is option B.
132 grams of propane would produce 6660 kJ of heat.
Step-by-step explanation:
Step-by-step explanation:

According to chemical reaction, combustion of 1 mole of propane gives 2,220 kilo Joules of heat , then 6,660 Kilo joules of heat will be obtained from :

Mass of 3.000 moles of propane :
3.000 mol × 44 g/mol = 132 g
132 grams of propane would produce 6660 kJ of heat.