NCEES 2001 #525

Can someone help explain how to get the solution for this problem. 

From the table below under the flue gas from stoic.,    ft3/ft3 fuel column,  read the moles of H2O produced from burning one mole of fuel.   (2) for methane(CH4), (3) for Ethane (4) for Propane.   Now since this a mixture of fuels multiply the %vol of each component.  So [(2)*(.859)+(3)*(.052)+(4)*(.015)] = 1.934  this is number of moles H2O produced from complete combustion.  Now multiply that by the lbm/mole of water.  2 moles of H = 2 lbm and 1 mole of O is 16lbm for a total of 18lbm per mole.  18*1.934 = 34.8 lbm.  

Thank you! A lot easier then I originally thought