What is the Identity of the Acid Used to Neutralize Aluminum Hydroxide?

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Discussion Overview

The discussion revolves around identifying the strong monoprotic acid used to neutralize aluminum hydroxide in a given chemical reaction. Participants explore stoichiometry, molar mass calculations, and the balancing of chemical equations as part of the problem-solving process.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the moles of aluminum hydroxide and expresses uncertainty about the mass of water used in the solution.
  • Another participant questions the stoichiometry of the reaction between the acid (HX) and aluminum hydroxide (Al(OH)3).
  • A participant suggests that the acid would have three moles based on the charge of aluminum.
  • There are references to a formula (m1v1 = m2v2) that some participants indicate may not be applicable to this problem.
  • One participant proposes balancing the reaction and using the masses to calculate the molar mass of the acid (HX).
  • A participant claims to have calculated the molar mass of HX to be approximately 100 g/mol and suggests that it could be HCLO4 based on this value.

Areas of Agreement / Disagreement

The discussion includes multiple viewpoints regarding the stoichiometry and balancing of the reaction, with no clear consensus on the identity of the acid or the correctness of the calculations presented.

Contextual Notes

Participants express uncertainty about the applicability of certain formulas and the accuracy of their calculations, indicating potential limitations in their approaches.

Who May Find This Useful

Students or individuals interested in chemistry, particularly those studying acid-base reactions and stoichiometry in a homework context.

Lori

Homework Statement


8.040 grams of a strong, monoprotic acid was added to enough water to make 500 ml of solution and used o completely neutralize 2.080 g aluminum hydroixide. identify the acid.

Homework Equations



m1v1 = m2v2
M = n/1 Liter

The Attempt at a Solution



i know that 2.08 g of Aluminum hydroxide is 0.02667 mols using molar mass of Aluminum hydroxide.
I also might know that there is 491.96 grams of water since i took the % from 8.040 grams/500 ml. Not sure if this is correct. I'm kinda stuck from here on.

I somewhat got a chemical equation going on with HX + Al(OH)3 --> H2O + AlX
 
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What do you think would be the stoichiometry of the HX - Al(OH)3 reaction?
 
You mean the mole conversions? I think X would have 3 mols right because Al has a +3 charge
 
Lori said:
m1v1 = m2v2

See my comment in your second titration thread. Forget you ever saw this formula.
 
Borek said:
See my comment in your second titration thread. Forget you ever saw this formula.
I'm aware that this formula won't work with this problem but I'm still stuck
 
Balance the reaction correctly and use the masses given to calculate molar mass of HX.
 
Ok. I got 100 grams/lol of HX using 3 moles HX/1mol al (oh)2

So Since it says strong acid, I just have to match it to the molar mass of the list of strong acids I know right? It would be HCLO4 since it also as like 100g/mol
 
Looks reasonable.
 

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