Molarity Calculation for Ba(NO3)2 Solution

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Discussion Overview

The discussion revolves around calculating the molarity of a barium nitrate solution formed from the reaction of barium hydroxide and nitric acid. It includes aspects of stoichiometry, limiting reagents, and the presence of excess reactants.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a balanced chemical equation for the reaction and attempts to calculate the moles of reactants involved.
  • Another participant suggests that the problem is a limiting reagent question and points out a potential typo in the formula for barium nitrate.
  • A participant confirms the correct formula for barium nitrate and encourages the use of the limiting reagent approach.
  • One participant calculates that 0.008 moles of barium nitrate will be formed and provides a molarity calculation based on the total volume of the solution.
  • Another participant notes that while the intermediate result of 0.008 moles of barium nitrate is correct, the participant has not addressed the excess reagent present in the solution.
  • A suggestion is made to list every ion and its concentration separately for clarity.
  • One participant expresses a lack of recent experience with limiting reagent problems, indicating a potential gap in understanding.

Areas of Agreement / Disagreement

Participants generally agree on the stoichiometry of the reaction and the calculation of moles, but there is disagreement regarding the completeness of the answer, particularly concerning the excess reagent and the need for detailed concentration listings.

Contextual Notes

There are unresolved aspects regarding the treatment of excess reagents and the final molarity calculation, which depend on the correct identification of all species in the solution.

Shackleford
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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
 
Last edited:
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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.
 
Why can I not edit my own damn post?
 
You can for 30 minutes, or something like that.
 
Borek said:
You can for 30 minutes, or something like that.

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

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

Now try the limiting reagent approach.
 
Borek said:
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
 
0.008 Ba(NO3)2 is a correct intermediate result. But that's not what you were asked about.
 
Borek said:
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.
 
  • #10
It is not all. Excess reagent is still present in the solution in unchanged form.
 
  • #11
Borek said:
It is not all. Excess reagent is still present in the solution in unchanged form.

Ah. I forgot about that.
 
Last edited:
  • #12
You should list every ion and its concentration separately. EOT for me.
 
  • #13
  • #14
End Of Thread, I believe. Do you understand the answer?
 

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