Answer:
The correct answer is option B.
132 grams of propane would produce 6660 kJ of heat.
Explanation:
Explanation:
[tex]C_3H_8+5O_2\rightarow 3CO_2+4H_2O,\Delta H=-2,220 kJ[/tex]
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 :
[tex]\frac{1}{-2,220}\times (-6,660 mol)=3.000 mol[/tex]
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.
