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

[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

 

[mjc123]

What do you think would be the stoichiometry of the HX - Al(OH)₃ reaction?

 

[Lori]

You mean the mole conversions? I think X would have 3 mols right because Al has a +3 charge

 

[Borek]

m1v1 = m2v2

See my comment in your second titration thread. Forget you ever saw this formula.

 

[Lori]

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

 

[Borek]

Balance the reaction correctly and use the masses given to calculate molar mass of HX.

 

[Lori]

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

 

[Borek]

Looks reasonable.