I looked up the heat of formation of AlCl3 and it is 704 kJ/mole; therefore, the delta H quoted in this problem is for the reaction (which is what it says in the problem but sometimes you can't believe what the problems say) for two moles.
Here's the hint.
If you had 54 grams Al (54/27 = 2 moles Al) how much heat would be produced? And you would tell me almost instantly the heat produced would be 1408.4 kJ and you would be right because the problem tells us that 2 moles Na will produce 1408.4 kJ.
Then I might ask a slightly different question. If you had 27 grams Al (27/27 = 1 mol Al) and you know 2 moles produces 1408.4 kJ) how much heat would we get. And you would answer almost instantly, "let's see now. If 2 moles produces 1408.4 kJ, then 1 mol must produce 1/2 of that (or if you want to do it with arithmetic it would be 1408.4 x 1/2 = 704.2 kJ)." And you would be right.
The final question is, if you had 4.10 g Al (4.10/27 = ?? moles) how many kJ of heat will you get. I think you see where we are going with this.
How much heat (in kilojoules) is released on reaction of 4.10 g of Al?
Reaction: 2Al(s) + 3Cl2(g)--->2AlCl3(s)
Delta H= -1408.4 KJ
-I converted grams to Al to moles
-and since there are 2 moles of Al, what do i do with the -1408.4 KJ?
please provide steps. thank you.
3 answers
wht is atom?
Questions that are piggy backed onto another question usually are overlooked. I suggest you go to the top of the page, click on post a new question, and type in your question.