BrO3- + 5Br- + 6H+ <--> 3Br2 + 3H2O

Question

If 25.0 ml of 0.200-molar BrO3- is mixed with 30.0 ml of 0.45-molar 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!

Answer 1

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) = ??

Answer 2

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.