Question:
A 5g sample of a hydrate of BaCl2 was heated, and only 4.3g of the anhydrous salt remained. What percentage of water was in the hydrate
Answer:
5g BaCl2 *xH2O ---> 4.3 g BaCl2 +0.7 g H2O
divide each by its molecular weight.
4.3/208= 0.02 mol BaCl2
0.7/18 = 0.04 mol H2O
Since the ratio of moles H2O to mole BaCl2 is 2, the dihydrate BaCl2*2H2O was created.
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