On heating 5.03 g of hydrated barium chloride (bacl2 xh2o), 4.23 g is left behind. what is the name of this hydrate? (

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

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