Question

What would be the formula of a barium iodide hydrate that contains `0.0243*mol` `BaI_2`, and `0.75*mol` water?

Answer 1

Well it would be `BaI_2*30H_2O`

Explanation:

And how did I get this?

I took the quotient `"Moles of water"/"Moles of barium iodide"`.

And thus we gets........`(0.75*mol*H_2O)/(0.0243*mol*BaI_2)`

I think that you have quoted the wrong units; barium iodide is typically supplied as the dihydrate, i.e. `BaI_2*2H_2O`; well, that's what the (old!) bottle on the shelf says.