A bomb calorimeter has a heat capacity of 2.47 kJ/K including the water. When a 0.111 g sample of ethylene (C 2H 4) was burned in this calorimeter, the temperature increased by 2.26 K. Calculate the energy of combustion for one mole of ethylene. a. -50.3 kJ/mol b. -1.41 x 103 kJ/mol c. -0.274 kJ/mol d. -624 kJ/mol e. -5.29 kJ/mol

Answer : The correct option is, (b)

Explanation :

First we have to calculate the heat produced.

where,

q = heat produced = ?

c = specific heat capacity =

= Change in temperature = 2.26 K

Now put all the given values in the above formula, we get:

Now we have to calculate the energy of combustion for one mole of ethylene.

where,

= energy of combustion for one mole of ethylene = ?

q = heat released = 5.5822 kJ

m = mass of = 0.111 g

Molar mass of = 28 g/mol

Now put all the given values in the above formula, we get:

Negative sign indicate the energy is released.

Therefore, the energy of combustion for one mole of ethylene is,

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g