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.
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