Molarity Calculation for Ba(NO3)2 Solution

[Shackleford]

Homework Statement

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.

Homework Equations

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

 

[Borek]

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.

 

[Shackleford]

Why can I not edit my own damn post?

 

[Borek]

You can for 30 minutes, or something like that.

 

[Shackleford]

You can for 30 minutes, or something like that.

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

 

[Borek]

Exactly.

Ba(OH)₂ + 2HNO₃ -> Ba(NO₃)₂ + 2H₂O

Now try the limiting reagent approach.

 

[Shackleford]

Exactly.

Ba(OH)₂ + 2HNO₃ -> Ba(NO₃)₂ + 2H₂O

Now try the limiting reagent approach.

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

 

[Borek]

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

 

[Shackleford]

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

Borek
--
equation balancer and stoichiometry calculator

I know. Check my previous post. I put the molarity of the solution.

 

[Borek]

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

 

[Shackleford]

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

Ah. I forgot about that.

 

[Borek]

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

 

[Shackleford]

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

Borek
--
equation balancer and stoichiometry calculator

I haven't done a limiting reagent problem in well over two years. EOT?

 

[chemisttree]

End Of Thread, I believe. Do you understand the answer?