2. When 0.7632 g of a hydrous ferric sulfate Fe2(SO4)3. n H2O salt was heated to remove all the water 0.5431 g of the anhydrous ferric sulfate (MW of Fe2(SO4)3 is 399.91) was obtained. What is the value of n in the hydrous salt

9

Explanation:

The following data were obtained from the question:

Mass of hydrous Fe2(SO4)3.nH2O = 0.7632 g

Mass of anhydrous Fe2(SO4)3 =

0.5431 g

Molecular weight of Fe2(SO4)3 = 399.91 g/mol

Value of n =..?

Next, we shall determine the mass of H2O in Fe2(SO4)3.nH2O. This can be obtained as follow:

Mass of hydrous Fe2(SO4)3.nH2O = 0.7632 g

Mass of anhydrous Fe2(SO4)3 =

0.5431 g

Mass of H2O =?

Mass of H2O = (Mass of hydrous Fe2(SO4)3.nH2O) – (Mass of anhydrous Fe2(SO4)3)

Mass of H2O = 0.7632 – 0.5431

Mass of H2O = 0.2201 g

Finally, we shall determine the value of n as follow:

Molar mass of Fe2(SO4)3.nH2O = 399.91 + n(2×1 + 16)

= 399.91 + 18n

Mass of Mass of hydrous Fe2(SO4)3.nH2O = 0.7632 g

Molar mass of nH2O = 18n

Mass of H2O = 0.2201 g

nH2o / Fe2(SO4)3.nH2O = mass of H2O / mass of Fe2(SO4)3.nH2O

18n/399.91 + 18n = 0.2201/0.7632

Cross multiply

18n × 0.7632 = 0.2201 (399.91 + 18n)

13.7376n = 88.020191 + 3.9618n

Collect like terms

13.7376n – 3.9618n = 88.020191

9.7758n = 88.020191

Divide both side by 9.7758

n = 88.020191 / 9.7758

n = 9

Thus, the value of n is 9

increased stability in the completion of fluorine's outer shell.