Using information in Appendices B and C, calculate the minimumnumber of grams of

Question

Using information in Appendices B and C, calculate the minimumnumber of grams of C3H8 (gas) that must be combusted to provide theenergy necessary to convert 1.95 kg of H20 from its solidform @ -13 C to its liquid form @ 80.0 C. Additional info/ assumptions/what I have done thus far: I assumed that the amount of heat required to heat the waterto 80 would equal the amount of heat given off by the combustion ofC3H8. I calculated the H needed in kJ to heat the water(1356.158 kJ) I assumed that I wanted to use the H offormation for C3H8 (-103.85 kJ/mol) and divide H required toheat the water by -103.85 kJ/mol. That would give me the # ofmol C3H3 required and I would convert that to grams, which wouldgive me my answer. I converted 1.95 kg H20 to 1950 g. In case I needed it, I converted it to moles as well. 1950g/18 g per mol=108.33 mol H20 I calculated the molecular weight of C3H8 to be 44.096g/mol 1. To raise the temp of the H20 from -13 to 0 I used theformula MCT, using the specific heat of water at Ice tobe 2.092 J/g-C. 1950g * 2.092 J/g-C * 13 = 53032.2 J 2. To account for the heat of fusion, I used 6.008 kJ/mol*108.33 mol = 650846.64 J 3. To raise the temp from 0 to 80 I used mcT again 1950 g * 4.184 J/g-C * 80= 652704 J 4. I added 53032.2 J + 650846.64 J + 652704 J = 135682.84J and divided that by 1000 to convert it to KJ =1356.58. I assumed that would be the amount of heat requiredto warm the H20 from -13 to 80. 5. Using -103.85 kJ/mol as the heat of fusion for C3H8 (gas) Iassumed dividing 1356.58/-103.85 would give me the # of molC3H8 I needed. I calculated 13.063 mol C3H8 required. 6. To convert mol C3H8 to grams I mulitiplied 13.063 mol * themolecuar weight of 44.096 =576.02 g required. There is a chance that this is correct. However, I used 2/3 chances w/a miscalculation in part 2 (wrong answer was 310g and a minor change in sig figs for 2 wrong answers ) on webassignand I do not want to use the last chance I have unless I'm prettysure I'm doing it correctly. And, this question hasn't beenanswered yet. Oh, and for part 2, I did calculate it using a g/mol heat offusion formula of 333.55 J/g, * 1950 but that gave me650422.5 J, which was 400+ off my answer, and since the questionspecifically said use the appendix in the back of the book, Idecided it was safer to convert the grams to moles and use 6.008kJ/mol. Please let me know if you see something wrong w/mycalculations, assumptions or answer. Using information in Appendices B and C, calculate the minimumnumber of grams of C3H8 (gas) that must be combusted to provide theenergy necessary to convert 1.95 kg of H20 from its solidform @ -13 C to its liquid form @ 80.0 C. Additional info/ assumptions/what I have done thus far: I assumed that the amount of heat required to heat the waterto 80 would equal the amount of heat given off by the combustion ofC3H8. I calculated the H needed in kJ to heat the water(1356.158 kJ) I assumed that I wanted to use the H offormation for C3H8 (-103.85 kJ/mol) and divide H required toheat the water by -103.85 kJ/mol. That would give me the # ofmol C3H3 required and I would convert that to grams, which wouldgive me my answer. I converted 1.95 kg H20 to 1950 g. In case I needed it, I converted it to moles as well. 1950g/18 g per mol=108.33 mol H20 I calculated the molecular weight of C3H8 to be 44.096g/mol 1. To raise the temp of the H20 from -13 to 0 I used theformula MCT, using the specific heat of water at Ice tobe 2.092 J/g-C. 1950g * 2.092 J/g-C * 13 = 53032.2 J 2. To account for the heat of fusion, I used 6.008 kJ/mol*108.33 mol = 650846.64 J 3. To raise the temp from 0 to 80 I used mcT again 1950 g * 4.184 J/g-C * 80= 652704 J 4. I added 53032.2 J + 650846.64 J + 652704 J = 135682.84J and divided that by 1000 to convert it to KJ =1356.58. I assumed that would be the amount of heat requiredto warm the H20 from -13 to 80. 5. Using -103.85 kJ/mol as the heat of fusion for C3H8 (gas) Iassumed dividing 1356.58/-103.85 would give me the # of molC3H8 I needed. I calculated 13.063 mol C3H8 required. 6. To convert mol C3H8 to grams I mulitiplied 13.063 mol * themolecuar weight of 44.096 =576.02 g required. There is a chance that this is correct. However, I used 2/3 chances w/a miscalculation in part 2 (wrong answer was 310g and a minor change in sig figs for 2 wrong answers ) on webassignand I do not want to use the last chance I have unless I'm prettysure I'm doing it correctly. And, this question hasn't beenanswered yet. Oh, and for part 2, I did calculate it using a g/mol heat offusion formula of 333.55 J/g, * 1950 but that gave me650422.5 J, which was 400+ off my answer, and since the questionspecifically said use the appendix in the back of the book, Idecided it was safer to convert the grams to moles and use 6.008kJ/mol. Please let me know if you see something wrong w/mycalculations, assumptions or answer.

Explanation / Answer

You should not have used the heat of fusion/formation for C3H8. Youshould have used the enthalpy of reaction of propane C3H8 + 5O2 => 3CO2  +   4H2O The delH_rxn is obtained from the enthalpy of formation differencebetween the products and reactants.

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