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On heating 5.03 g of hydrated barium chloride (BaCl2 XH2O), 4.23 g is left behind. What is the name of this hydrate? (

User Vencat
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1 Answer

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1) Deduce the two masses and see the amount of water was driven off when heated:

5.03 g - 4.23 g = 0.8 g H2O given off

2) Change mass from grams to moles of H2O:

0.8 g H2O / 18 g H2O in 1 mole = 0.044 mol H2O

3) Change left over mass to moles of BaCl2 .


4.23 g BaCl2 / 207 g BaCl2 in 1 mol = 0.021 mol BaCl2

4)Find the ratio of mol H2O to mol BaCl2:

0.044 mol H2O : 0.021 mol BaCl2

5) The resulting ratio is 2:1 so two H2O for each BaCl2, thus, the hydrate was named:
Barium chloride di-hydrate

User Lky
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