chem help please
posted by janice on .
If 720 mL of 0.0165 M aqueous Al3+ and 300 mL of 0.0567 M aqueous Ca2+ are reacted, how many mol of Br- is produced?
Al2(SO4)3(aq) + 3 CaBr2(aq) → 2 AlBr3(aq) + 3 CaSO4(s)
my answer was .03564 mol of Br will be produced but the real answer is .0340 mol...i do not know what i did incorrectly
This looks like a limiting reactant question. How many moles of each reactant..
Al2(SO4)3 moles .720*.0165=.01188
Hmmm. The balaced equation requires three times as much calcium bromide as aluminum sulfate.
So, the calcium bromide is the limiting reageant.
Moles of AlBr3 produced= 2/3 (.0170 and moles of Br- involved is 3*2/3*.0170 or .0340 moles.
I would like to point out a flaw in the question. If no matter how much Al^+3 reacts with no matter how much Ca^+2, then NO Br^- will be produced. You don't have any Br^- there to begin with.