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A sample of ethanol, C2H5OH, weighing 2.84 g was burned in an excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rose from 25.00°C to 33.73°C. If the heat capacity of the calorimeter and contents was 9.63 kJ/°C, what is the value of q for burning 1.28 mol of ethanol at constant volume and 25.00°C?
The reaction is shown below.
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

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asked by Lance

  1. q for calorimeter is heat cap x delta T.
    qcal = 9.63 kJ/C x (33.73-25.00) = 84.1 kJ and that is for 2.84 g ethanol. I would convert 2.84 g ethanol to moles.
    2.84/46 = 0.0617 moles; therefore, the heat of burning is 84.1 kJ/0.0617 = 1,362 kJ/mol. Now all you need to do is determine the amount of heat from burning 1.28 mol since you have the heat for 1 mole. Check my work above and round everything to the appropriate number of significant figures. Also, I estimated the molar mass of C2H5OH. It may not be exactly 46.0.

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    posted by DrBob222

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