I took the average and got 23.30ml = 0.0233 L
This is just the 0.0987 M * 0.0233L = 0.00230 mols right?
So there are half the mols of Ba(OH)2 as HCl...giving me 0.00230/2 = 0.00115 mols Ba(OH)2 and 0.19704 grams.
So then the difference is simply 3.632 g - 0.19704 g = 3.435 g H20...
Barium hydroxide forms several hydrates. A specimen of barium hydroxide, suspected of being a hydrate, was prepared and analyzed as follows to determine its formula.
3.632 g of the compound was dissolved in water to give 250.0mL of solution. 25.00 mL of this solution was titrated with 0.0987...