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A hydrocarbon fuel of average composition of eicosene is burned in a furnace at 1atm pressure in such a way that the combustion gas analysis is 11% of CO2 AND 0.5% of CO. The air and fuel enter the furnance at 15 degree celsius and leaved at 204.4 degree celcius. Taking into account stack losses,what is the amount of heat available for transfer to the furnace per kg of fuel burned? The heat of combustion of the liquid fuel at 15 degree celcius and 1 atm is 45261.233kj/kg
Weegy: First balance the equation C3H8 + 5O2 - - - - - - - > 3CO2 + 4H2O Convert the mass of C3H8 to moles... [ [ Always work with moles moles = mass / molar mass molar mass C3H8 = (3 x 12.01) + (8 x 1.008) = 44.094 g/mol moles C3H8 = 5.53 g / 44.094 g/mol = 0.1254 moles C3H8 You need enough moles of O2 to react 0.1254 moles of C3H8, so work out how many moles of O2 are needed to react with 0.1254 moles C3H8. The balanced equation tells you that 1 mole C3H8 needs 5 moles O2 to fully react So 0.1254 moles C3H8 needs (5 x 0.1254) moles O2 = 0.6270 moles O2 required to react all the propane. You need 0.6270 moles of O2 Now use PV = nRT to determine the volume of O2 at 25 deg C and 1.04 atm that has 0.627 moles of O2 P = pressure (1.04 atm) V = volume (unknown - Litres) n = moles (0.627 mol) T = Temp, must be in Kelvin. Kelvin = deg C + 273.15 so T = 298.15 K (25 deg C = (25 + 273.15 = 298.15 K)) R = gas constant (different for different units of Pressure and volume - see link) For P in atm and Vol in Litres R = 0.0821 L atm K^-1 mol^-1 V = nRT / P V = (0.627 mol x 298.15 K x 0.0821L atm K^-1 mol^-1) / 1.04 atm V = 14.8 L (3 sig figs) ] ] (More)
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