LP gas burns according to the exothermic reaction:

Question

C3H8(g) + 5 O2(g) right arrow 3 CO2(g) + 4 H2O(g) ΔHrxn° = −2044 kJ
What mass of LP gas is necessary to heat 1.2 L of water from room temperature (25.0°C) to boiling (100.0°C)? Assume that during heating, 18% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.

Answer 1

My calculator is broken; I'll try to describe this. First, since the LP gas is only 18% efficient, we may as well adjust that early.
2044 kJ(@100%) x 0.18 = about 400 kJ at 18%.
q needed = mass H2O x specific heat H2O x (Tfinal-Tinitial) = estimated 360,000 J.
So 44 g C3H8(1 mol) will produce about 400 kJ at 18% efficiency. What amount will produce 360 kJ?
44 g x (360/400) = ? g LP gas.