A #"140.5-g"# sample of #"NiSO"_4 * x "H"_2"O"# was heated until no further decrease in mass was observed. The mass of the anhydrous (dry) salt is #"77.5 g"#. Determine the value of #x# in the formula?

Question

20091 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-26T13:01:47

Check the answer below...

enter image source here

Answer 2 · 2018-02-26T14:04:27

#x=7#....

We start with the HYDRATE of a nickel salt...

#NiSO_4*n(H_2O)(s)+Delta rarr NiSO_4(s) + nH_2O(g)uarr#

We heat the nickel hydrate such that we are left with nickel sulfate, and the mass loss represents the moles of water...

#"Moles of nickel sulfate (anhydrous)"=(77.5*g)/(154.75*g*mol^-1)=0.501*mol#...

But the mass LOST represented the water present....

#"Moles of water lost"=(140.5*g-77.5*g)/(18.01*g*mol^-1)=3.50*mol#...and so we divide thru by the LOWEST molar quantity to get...

#(NiSO_4)_((0.501*mol)/(0.501*mol))*(OH_2)_((3.50*mol)/(0.501*mol))-=(NiSO_4)*(OH_2)_7#..

And so we got...#NiSO_4*(OH_2)_7#...a salt of nickel sulfate with SEVEN waters of crystallization....