One of the hydrates of Ba(ClO4)2 is barium perchlorate trihydrate. A 59.7 gram sample of Ba(ClO4)2...
Homework Answers
Mass of Ba(ClO4)2.3H2O = 59.7 g
moles of Ba(ClO4)2.3H2O = (mass of Ba(ClO4)2.3H2O) / (molar mass of Ba(ClO4)2.3H2O)
moles of Ba(ClO4)2.3H2O = (59.7 g) / (390.274 g/mol)
moles of Ba(ClO4)2.3H2O = 0.153 mol
After heating, water of hydration evaporates and anhydrous salt is left
moles of Ba(ClO4)2 = moles of moles of Ba(ClO4)2.3H2O
moles of Ba(ClO4)2 = 0.153 moles
mass of Ba(ClO4)2 = (moles of Ba(ClO4)2) * (molar mass of Ba(ClO4)2)
mass of Ba(ClO4)2 = (0.153 mol) * (336.2282 g/mol)
mass of Ba(ClO4)2 = 51.4 g
After heating, 51.4 grams of anhydrous compound remained
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One
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Stoichiometry
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