Question

A `0.49*g` mass of water is LOST from a mass of `3.75*g` `Fe(NO_3)_3*2H_2O` upon prolonged heating. What is the percentage composition of water with respect to the starting salt?

Answer 1

Approx. `3*g`..........

Explanation:

You gots `Fe(NO_3)_3*2H_2O`, which has a formula mass of `277.88*g*mol^-1`.

And so you got a molar quantity of `(3.75*g)/(277.88*g*mol^-1)=0.0135*mol`......

ANd now you subject this to drying to remove the water.....

`Fe(NO_3)_3*2H_2O+Delta rarr Fe(NO_3)_3(s) + 2H_2O(g)uarr`

And now you still have a `0.0135` molar quantity WITH RESPECT to `Fe(NO_3)_3`, i.e. `0.0135*molxx241.86*g*mol^-1=3.26*g`...

And so in fact you have lost a mass of `0.49*g` with respect to water....which is necessarily (why?) TWICE the molar quantity of starting hydrate salt.....

Finally we address the percentage of water by mass in the `"iron(II) hydrate"`, and this is simpy `"mass of water"/"mass of hydrate salt"xx100%`...

`=(0.49*g)/(3.75*g)xx100%=13.1%`. Agreed?