Ethanol combustion reaction is given below. CH3CHZOH() 302(g) 2C02(g) 3H2O(g) AH = - 1236 kJ a) If 15.3 grams of C2H5OH is converted to products, how much heat; in kJ and in kcal, is absorbed?

Question

calculate the number of moles in 15.3 grams of ethanol

molar mass ethanol = 46.07 g/mol

moles ethanol = mass ethanol/molar mass of ethanol

moles = 15.3 / 46.07 = 0.3321 moles ethanol

we know that 1-mole releases - 1236 KJ

Standard enthalpy change of combustion, ΔH°

The standard enthalpy change of combustion of a compound is the enthalpy change  (delta H) which occurs when one mole of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

0.3321 moles ethanol will release 0.3321 moles ethanol* -1236 kJ =  -410.4756 kJ

we know that 1 kcal = 4.184 kj so; - 410.4756 KJ * 1 kcal / 4.184 = - 98.1 Kcal

b. According to the statement 3 moles of water are produced in order to release 1236 KJ

find the moles in 42.7 grams of water

molar mass water = 18.015 g/mol

moles = mass / molar mass = 42.7 / 18.015 = 2.37 moles of water

heat released in k = 2.37 moles of water * -1236 KJ / 3 moles water 

= -976.44 KJ, this is the amount of heat released, the negative sign indicates heat release

Answer 1

Using the mass of the given ethanol (C₂H₅OH), use dimensional analysis to find how much heat is released.

First convert to moles using molar mass.

Second, using the enthalpy (Delta H) and corresponding moles of ethanol.

15.3 g C₂H₅OH x 1 mol C₂H₅OH x -1236 kJ = -410. kJ

46.08 g C₂H₅OH. 1 mol C₂H₅OH

The combustion releases 410. kJ of energy which can be absorbed.

Now convert.

410 kJ x 1 kcal/4.184 kJ = 98.1 kcal