[azurken]

Carry out a material balance for the reaction that occurs when 18.25 mL of 0.252 M
HBr is combined with 0.150 g of Zn(OH)2 solid.

Attempt:

So I started off with looking for the equation.

And came up with 2HBr (aq) + Zn(OH)2 (s) -> 2H2O (L) + Zn (aq) + 2Br (aq)

I know that Zn and Br is soluble with each other but there will also be leftover H ions.

Material balancing:

I start with

4.599 mmols of HBr
1.509 mmols of Zn(OH)2.

This translates to:

3.018 mmols of OH
4.599 mmols of H
4.599 mmols of Br
1.509 mmols of Zn

Then the final products:

1.581 mmol of H leftover              0.0866 M H
3.018 mmol of H2O formed            (am I supposed to be calculating this since its actually the liquid or do I skip it?)
1.509 mmol Zn unused                 0.0827 M Zn 
4.599 mmol Br unused                  0.252 M Br

Am I supposed to count the water in the end? and are there any other mistakes I've made?

[Hunter2]

No,

You calculate the moles of HBr and Zn(OH)₂

You got already 4.599 mmol HBr and 1.509 mmol Zn(OH)₂

In the equation you can see 2 HBr react with one Zn(OH)₂

For 4,599 mmol HBr you need 2.2995 mmol Zn(OH)₂, but you have only 1.509 mmol

So only (1.509 x 2 mmol) = 3.018 mmol HBr will react. 1.581 mmol will be left