Question

Given that the heat of combustion of propane, C₃H₈, has been determined to be −103.85 kJ/mol, predict the heat change when 100.0 grams of propane is burned in a combustion reaction (assuming no work done).

		q = −42.47 kJ
		q = −1039 kJ
		q = −45.80 kJ
		q = −235.5 kJ
		q = −352.5 kJ

Answer 1

molar mass of propane = 44 g/mol
number of mole of propane = (given mass)/(molar mass)
= 100/44
= 2.27 mole

reaction taking place is
C3H8 + 5O2 --> 3CO2 + 4H2O
heat of combustion of propane is −103.85 kJ/mol
which mean
1 mole of propane give −103.85 kJ of heat
so,
2.27 mole of propane give (−103.85*2.27) kJ of heat
2.27 mole of propane give -235.7 kJ of heat
heat change when 100.0 grams of propane is burned = -235.7 KJ

nearly equal to -235.5 KJ

Answer : -235.5 KJ