[bananafish]

OK, Aluminum and Bromine Br2 react to form AlBr3. What mass of AlBr3 will be produced if 1.00 gram of Al completely reacts with excess Br2? I've just started Chemistry and am a little confused with how to set it up. I know so far the way I have been doing it is wrong because I have the final answer from my study group, I just can't figure out the calculations. Any pointers would be great! Thanks a lot

[Borek]

This is a simple stoichiometry. Start with the balanced reaction equation. Do you know how to read it?

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations

[bananafish]

Ok, thanks! That website really helped. So, I attempted the balance to balance the reaction. Sorry, I am so new at this. 

I did 3Albr2--->2AlBr3

Now I am going to use the Mole Method, but I seem to get a little confused about setting up my equation.

1.00 g AlBr3            2 mol AlBr3                3 mol AlBr2              186.78 g AlBr2
___________    x   _______________  x ________________  x _________________
      1                      266.68 g AlBr3           2 mol AlBr3             1 mol



Am I on the right track at all? Can you tell me what I am doing wrong setting up my equation? Many thanks!

[Borek]

There is no such thing as AlBr₂, you react Al with Br₂, so your reaction starts as

Al + Br₂ -> ...

[bananafish]

2Al + 3Br2 ----> 2AlBr3

OK, can you help me out with where to go here? I have read the ChemBuddy link you sent a couple of times and am not sure whether to use Dimensional Analysis or the Mole Ratio technique. Basically both of them I getting really turned around. I know I am looking for what mass will be produced of AlBr3 but I just get lost setting it up initially.

[bananafish]

I have now done

1.00 g Al        1 mol Al           3 mol Br
_________  x   ________    x  ________
   1                  26.98 g Al      2 mol Al


now unsure where to go from here.

[fledarmus]

Think about what the balanced equation is telling you. In words, it says the 2 moles of Al will react with 3 moles of Br2 to form 2 moles of AlBr3. The first conversion factor you have in place (1 mol Al/26.98 g Al) converts the grams of aluminum to the moles of aluminum. Your second factor (3 mol Br/2 mol Al) is meaningless - there is no Br in your reaction. Instead, you should be converting the moles of Al consumed to the moles of AlBr3 formed.