# Homework Help: Easy CHEM problem

1. Jul 29, 2008

### Shackleford

1. The problem statement, all variables and given/known data

Find the molarity of the solution (salt and any excess reagent after reaction of 250 mL of 0.05 M Ba(OH)2 with 200 mL of 0.08 M HNO3.

2. Relevant equations

3. The attempt at a solution

I am drawing a blank. Here's the balanced equation

Ba(OH)2 + 2HNO3 ===> 2H2O + Ba(NO3)s

Ba(OH) .0125 moles

2HNO3 .016 moles

Last edited: Jul 29, 2008
2. Jul 30, 2008

### Staff: Mentor

Simple stoichiometry, this is a limiting reagent question. Your reaction equation is almost OK (chack barium nitrate formula, but I suppose that's a typo) and is a correct first step to solution.

3. Jul 30, 2008

### Shackleford

Why can I not edit my own damn post?

4. Jul 30, 2008

### Staff: Mentor

You can for 30 minutes, or something like that.

5. Jul 30, 2008

### Shackleford

Ok, well, Ba(NO3)2, correct?

6. Jul 30, 2008

### Staff: Mentor

Exactly.

Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O

Now try the limiting reagent approach.

7. Jul 30, 2008

### Shackleford

I used the number of moles not grams to find the limiting reagent, which is nitric acid. Only .008 moles of barium nitrate will be formed.

.008 moles/.450 L = .017 M

8. Jul 30, 2008

### Staff: Mentor

0.008 Ba(NO3)2 is a correct intermediate result. But that's not what you were asked about.

9. Jul 30, 2008

### Shackleford

10. Jul 30, 2008

### Staff: Mentor

It is not all. Excess reagent is still present in the solution in unchanged form.

11. Jul 30, 2008

### Shackleford

Last edited: Jul 30, 2008
12. Jul 30, 2008

### Staff: Mentor

You should list every ion and its concentration separately. EOT for me.

13. Jul 30, 2008

### Shackleford

14. Jul 31, 2008