Chemistry
posted by Andrew .
BrO3 + 5Br + 6H+ <> 3Br2 + 3H2O
If 25.0 ml of 0.200molar BrO3 is mixed with 30.0 ml of 0.45molar Br solution that contains a large excess of H+, what is the amount of Br2 formed, according to the above equation?
The answer is 0.00810 moles but I don't know how to approach this!
I have the final exam tomorrow. Pleas help me!
Thanks!

First, recognize that this is a limiting reagent problem. You can tell that when BOTH M and L of BOTH reactants are given.
moles BrO3^ = M x L = 0.200 x 0.025 = 0.00500 (This one; i.e., M x L will get moles every time.)
Now to find mols Br2 formed, we convert moles BrO3^ to mols Br2 using the coefficient in the balanced equation (dimensional analysis).
So we would have 0.015 mols Br2 IF we had all the Br^ we need to react with all of the bromate ion.
Now we do the same thing with Br^.
moles Br^ = M x L = 0.45 x 0.030 = 0.0135 moles bromide ion.
Convert to moles Br2 using the equation.
That will be
0.0135 moles Br x (3/5) = 0.00810 moles.
The smaller number means that one is the limiting reagent; therefore, 0.00810 is the answer to the problem. Check my work and check for typos.
moles Br2 = moles BrO3^ x (3 moles Br2/1 mol BrO3^) =
0.00500 x (3/1) = ?? 
The explanation for the first part didn't copy so here it is again.
Now to find mols Br2 formed, we convert moles BrO3^ to moles Br2 using the coefficients in the balanced equation (dimensional analysis).
So we would have 0.015 mols Br2 IF we had all the Br^ we need to react with all of the bromate ion.
moles Br2 = moles BrO3^ x (3 moles Br2/1 mol BrO3^) = 0.0050 x (3/1) = 0.015 moles Br2 IF we had all the Br^ we needed to react with ALL of the bromate ion.