One of the hydrates of MnSO4 is manganese (II) sulfate tetrahydrate. A 71.6 gram sample of MnSO4 4 H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?

48.32 g of anhydrous MnSO4.

Explanation:

Equation of dehydration reaction:

MnSO4 •4H2O - -> MnSO4 + 4H2O

Molar mass = 55 + 32 + (4*16) + 4 ((1*2) + 16)

= 223 g/mol

Mass of MnSO4 • 4H2O = 71.6 g

Number of moles = mass/molar mass

= 71.6/223

= 0.32 mol.

By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4

Number of moles of MnSO4 = 0.32 mol.

Molar mass = 55 + 32 + (4*16)

= 151 g/mol.

Mass = 151 * 0.32

= 48.32 g of anhydrous MnSO4.