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 propane, C3H8(g),that must be combusted to provide the energy necessary to convert3.33 kg of ice at -13.0°C to liquid waterat 75.0°C.
The Specific heat..   ice is 2.092 j/g K the specific heat.... of water is 4.184 j/gK
delta H of ice is (6.008 Kj/ mol) Appedecies C H2O (l) is -285.83 kj/mol   CO2 (g) is -393.5 kj/mol C3H8 is -103.85 i did the heating curve and the the formulas q=mc deltaT and added all the Q's toghether then what do i do now pleasehelp Using information in Appendices B and C, calculate the minimumnumber of grams of propane, C3H8(g),that must be combusted to provide the energy necessary to convert3.33 kg of ice at -13.0°C to liquid waterat 75.0°C.
The Specific heat..   ice is 2.092 j/g K the specific heat.... of water is 4.184 j/gK
delta H of ice is (6.008 Kj/ mol) Appedecies C H2O (l) is -285.83 kj/mol   CO2 (g) is -393.5 kj/mol C3H8 is -103.85 i did the heating curve and the the formulas q=mc deltaT and added all the Q's toghether then what do i do now pleasehelp

Explanation / Answer

After adding all the Q's ; divide the sum of Q by theamount of heat of combutions of C3H8 per mole to get the number of moles of C3H8 required. C3H8 (g) +5 O2 (g).............> 3CO2 (g) + 4H2O (l) ;calculate H0rxn for this reaction which will giveyou heat of combution of C3H8. i,e number of moles of C3H8 required = QkJ/ heat ofcombution per mole of C3H8 The above equation will give you an idea.

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