What mass of LP gas is necessary to heat 1.8 of water from room temperature (25.0 ) to boiling (100.0 )?

Question

What mass of LP gas is necessary to heat 1.8  of water from room temperature (25.0 ) to boiling (100.0 )?

 Liquefied petroleum (LP) gas burns according to the following exothermic reaction: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) ΔH∘rxn=−2044kJ

Answer 1

First, find out how much heat (in joules or kjoules) you need to heat the water:

q = mC∆T

q = heat = ?

m = mass = 1.8 kg (assuming this is the case since you didn't put any units for 1.8 of water)

C = specific heat = 4.184 J/gº = 4.184 kJ/kgº

∆T = change in temperature = 75º (100º - 25º)

q = (1.8 kg)(4.184 kJ/kgº)(75º)

q = 564.8 kJ of heat needed

Next, find out how many grams of LP are needed to provide this much heat:

LP = liquid propane (C3H8)

1 mol LP = 2044 kJ of heat according to the balanced equation

564.8 kJ x 1 mol LP / 2044 kJ = 0.276 mols LP needed

molar mass LP = 44.1 g / mol

mass of LP necessary = 0.276 mols x 44.1 g /mol = 12.2 g LP