What is the hydrate formula for something that is .0243 mol #BaI_2# and .098 mol #H_2O#?

1 Answer
Apr 21, 2016

Answer:

#"BaI"_2 * 4"H"_2"O"#

Explanation:

The idea here is that you need to use the number of moles of anhydrous barium iodide, #"BaI"_2#, and of water, #"H"_2"O"#, to find the mole ratio that exists between the anhydrous salt and its water of hydration in the hydrate.

So, divide both values by the smallest one to find

#"For BaI"_2: color(white)(a)(0.0243 color(red)(cancel(color(black)("moles"))))/(0.0243color(red)(cancel(color(black)("moles")))) = 1#

#"For H"_2"O: " (0.098 color(red)(cancel(color(black)("moles"))))/(0.0243color(red)(cancel(color(black)("moles")))) = 4.033 ~~ 4#

This means that one formula unit of this hydrate will contain #1# mole of barium iodide and #4# moles of water. Its chemical formula will thus be

#color(green)(|bar(ul(color(white)(a/a)color(black)("BaI"_2 * 4"H"_2"O")color(white)(a/a)|))) -># barium iodide tetrahydrate