Tried this question and got 32,000J as the answer, but it says its incorrect; an

Question

Tried this question and got 32,000J as the answer, but it says its incorrect; any in depth help would be greatly appreciated!

Thanks!

From the data below, calculate the total heat (in J) needed to convert 0.504 mol of gaseous ethanol at 300.0 C and 1 atm to liquid ethanol at 25.0°C and 1 atm: gaseous ethanol at 300.0° C and 1 atm to liquid ethanol at 25.0° C and 1 atm vap 40.5 kJ/mol Cethanoll): 2.45J/g. °o b.p. at l atm : 78.5°C AH rap: 40.5 H Cethanolg) ethanol(g): 1.43 J/g·"C 1.43J/g- *c cethanoll): 2.4

Explanation / Answer

Molar Mass of Ethanol (C2H5OH) = 2 * 12 + 6 * 1 + 1 * 16 = 46 gm/mol

Mass of C2H5OH = 0.504 * 46 = 23.184 gms

The heat required to cool down it from the temperature of 300C to 78.5C is equal to

mass * specific heat * (temp change)

=> 23.184 * 1.43 * (78.5-300)

=> -7343.4 J

Heat required for condensation = number of moles * heat vaporization

=> 0.504 * (-40.5 KJ)

=> -20160 J

heat required to cool down to 25C

=> 23.184 * 2.45 * (25-78.5)

=> -3038.84

Total Answer = -7343.4 - 3038.84 - 20160

=> -30542

Hence the total heat needed is equal to 30542J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.