PLEASE HELP! IM STUMPED :( How many kilograms of propane would be needed to raise the temperature of the air?

Question

A house with a 16.2 × 103 ft2 floor area and 8.50 ft high ceilings is to be heated with propane. How many kilograms of propane would be needed to raise the temperature of the air inside the house from 63.0 oF to
73.0 oF ?. In this calculation your are ignoring the temperature of the walls etc.



The moloar specific heat of the air inside the house is 30.2 J mol −1 oC −1
The average density of air over this temperature range can be estimated to be 1.22 g/L.
The molar mass of air is 29.1 g/mol.
There are 28.317 liters in a cubic foot.
The molar heat of combustion of propane is −2220.047 kJ/mol


PLEAASE HELPPP I NEED ANSWER + EXPLANATION IF POSSIBLE :) thankyoooou!

Answer 1

Volume of house is (16.2 x 10^3) x 8.5 = 138,000 cubic feet
Mass of air in house = 138,000 ft3 x 28.317 l/ft3 x 1.22 g/l = 4,770,000 grams
4,770,000 grams / 29.1 g/mol = 164,000 moles of air in house
Temperature rise = 73F - 63F = 10F = 5.56C
Heat required = 164,000 mol x 5.56C x 30.2 J/mol-C = 27,500,000J = 27,500 kJ
Burning propane generates 2,220.047 kJ/mol, so 27,500 kJ would require the combustion of 27,500 / 2,220 = 12.4 moles propane (assuming 100% efficiency)
Molecular weight of propane (C3H8) is 12 x 3 + 1 x 8 = 44 g/mol
12.4 moles propane x 44 g/mol = 546 g propane = 0.546 kg

Not much, is it?