calculate the minimum number of grams of propane, C 3 H 8 thatmust be combusted
ID: 690393 • Letter: C
Question
calculate the minimum number of grams of propane, C3H8 thatmust be combusted to provide the energy necessary to convert 4.85kg of ice at -20.5 degrees celcius to liquid water at 74.0 degresscelciius. I know how to calculate the energy need to reach the 80degrees. When setting up the balanced equation i got: C3H8(g)+5H2O(g) yields3H2O(g) +4CO2(g) I used enthalpies of formation but I didnt recieve thecorrect answer. I am thinking that the enthalpy of formation forH2Oshould be of the liquid phase? calculate the minimum number of grams of propane, C3H8 thatmust be combusted to provide the energy necessary to convert 4.85kg of ice at -20.5 degrees celcius to liquid water at 74.0 degresscelciius. I know how to calculate the energy need to reach the 80degrees. When setting up the balanced equation i got: C3H8(g)+5H2O(g) yields3H2O(g) +4CO2(g) I used enthalpies of formation but I didnt recieve thecorrect answer. I am thinking that the enthalpy of formation forH2Oshould be of the liquid phase? I used enthalpies of formation but I didnt recieve thecorrect answer. I am thinking that the enthalpy of formation forH2Oshould be of the liquid phase?Explanation / Answer
The balanced equation is: C3H8(g)+5 O2(g)-----> 4H2O(g) +3CO2(g) When propane burns it gives out water vapor. So H2O (g) iscorrect The heat of reaction, Hf = { (4 (-241.8 kJ/mol) + 3 (-393.5 kJ/mol) } - { (-103.8kJ/ml) + 5 (0) } = - 2147.7 + 103.8 = 2044 kJ This is the heat given out by combustion of one mole ofpropane. . Energy required by ice: total energy required = Ice at -20.5 to ice at 0 + Ice at 0 towater at 0 + water at 0 to water at 74.0 C = 4.85 *10^3 g * 2.05 J/g.C *(20.5-0)C + 4.85 * 10^3 g * 333.55 J/g+4.85 *10^3g * 4.18 J/g.C * (74.0-0) = 684.89 * 4.85 * 10^3 J = 3322 kJ . Moles of propane required = 3322 kJ * (1 mole / 2044 kJ) = 1.63 moles . mass of propane required = moles * molar mass =1.63 moles * 44.10 g/mol =71.9 g
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