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

1 Answer
Jun 10, 2017

Answer:

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.