A 55.0-mg sample of Al(OH)3 is reacted with 0.200M HCl. How many milliters of the acid are needed to neutralize the Al(OH)3.

2 answers

Emily, this is worked almost the same way as the NaOH/H2SO4 problem I worked for you earlier. What is it you don't understand about this?
All of them use this standard format.
1. Convert from what you have to mols.
2. Convert mols of what you have to mols of what you want.
3. Convert mols of what you want to whatever units you want; i.e., grams, liters, etc.
The equation goes like this. (Un-labeled)
0.055g x (1 mol)/(78g) x (3 mol)/(1 mol) x (1L)/(0.200M) = 0.01057
You then convert this to mL
0.01057 * 1000 = (10.6 mL HCL) <-- rounded to sf
Hope this helps! (10 years later) lol