What Happens When Aluminium Reacts with Chlorine in Water?

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ShakeSpee
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Hi, thanks for checking out this thread.

I'm really having a problem with the question because I don't have any foundation.

If you could, would you:
Explain what exactly is going on? Is this an endothermic or exothermic reaction?
What are colligative properties, and what do they have to do with this question?

Homework Statement



10.0 g Aluminium and 21.0 g Chlorine react in a solution of water. Ignoring colligative properties, what will the state of water be if 1000.0g starts at 10.0°C and a heat of formation of 92kJ/mol of product?

Thanks again for taking the time to help.
--ShakeSpee
 
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Okay, so far I think I balanced the equation (but I don't think I did it correctly.)
1 Al(s) + 3 Cl(l) -----> 1AlCl3 +H20 + energy

I also calculate the amount of moles. (.371 mol Al, 1.777 mol Cl)

Is this correct?
 
Now I'm stuck; do I have to find the change in temperature to find the state of water?
 
ShakeSpee said:
Okay, so far I think I balanced the equation (but I don't think I did it correctly.)
1 Al(s) + 3 Cl(l) -----> 1AlCl3 +H20 + energy

Chlorine is diatomic (not that it will change the result in this case).

I also calculate the amount of moles. (.371 mol Al, 1.777 mol Cl)

Al is OK, Cl is not (regardless of whether it is to be treated as atomic, or diatomic).

You need to find the limiting reagent and calculate number of moles of AlCl3 produced. Then you will be able to calculate amount of heat produced, and use it to calculate new water temperature.
 
Borek said:
Chlorine is diatomic (not that it will change the result in this case).
Al is OK, Cl is not (regardless of whether it is to be treated as atomic, or diatomic).

You need to find the limiting reagent and calculate number of moles of AlCl3 produced. Then you will be able to calculate amount of heat produced, and use it to calculate new water temperature.
I thought that was where my mistake was! I try it again, thanks!