Question

One of the hydrates of BaI2 is barium iodide hexahydate. A 88.3 gram sample of BaI26H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained? ________ g

Answer 1

molar mass of BaI2 = 391.13 g / mol

molar mass of BaI2 6H2O = 499.25 g / mol

moles of BaI2 6H2O = 88.3 / 499.25

                                 = 0.177

moles of BaI2 = 0.177

mass of BaI2 = moles x molar mass

                     = 0.177 x 391.13

                    = 69.2 g

the anhydrous compound remained = 69.2 g