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?

1 Answer
anor277
Sep 16, 2017

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?

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