Enthalpy question

Question

The combustion of propane is given by the following reaction.

C₃H₈ + 3 O₂ → 3 CO₂ + 4 H₂O

The enthalpy of reaction is −2202.0 ^kJ/_mol. How much energy (in joules) will be released if 47.99 grams of propane is burned. (Molar mass of propane = 44.11 g/mol). 

Answer 1

The key here is following your units. They are asking you to report the answer in Joules and you have been given everything you need in order to be able to do so.

You have also been given a balanced equation:

C₃H₈ + 3 O₂ → 3 CO₂ + 4 H₂O

You will use dimensional analysis until you achieve the units Joules.

You have been given the exact amount of Kj for the reaction for 1 mole. This is -2202.0kJ/mol.

You are also given the clue that propane is burned with this amount of grams: 47.99 g. This is simply just stating that this is a combustion reaction and identifying that this is the amount of grams related to this reaction.

Take your grams for proprane and convert it to moles using its atomic weight given in problem of 44.11 g/mol.

So you can have moles you will write it 1mol/44.11g x 47.99g this way grams cancel grams top to bottom and you will be left with moles in the numerator.

The answer will be 1 multiplying it by 47.99 and then dividing by 44.11:

(1 x 47.99)

-------------- =

44.11

1.08796 moles

Now that you have moles, take it to kJ or KiloJoules

1.08796mol x -2202.0kJ/mol moles will cancel moles top to bottom and you will be left with kJ:

-2395.69kJ

now take it to Joules

1000 J= 1kJ

1000 J/1 kJ x -2395.69kJ kJ cancel kJ top to bottom and you are left with J or Joules which is what they are looking for so you should have:

-2395692.13 J is what is related to this reaction. Combustion reactions give off a lot of energy especially a lot as heat and this makes sense.

For sig fig purposes you can report it to 4 since its the least you have been given so:

-2396000 J