After 0.6523g of CoCl 2 *6H 2 O is heated,the residue has a mass of 0.4323g. Cal

Question

After 0.6523g of CoCl2*6H2O is heated,the residue has a mass of 0.4323g. Calculate the % H2Oin the hydrate. What is the actual number of moles of water performula unit CoCl2? I don't even know where to start! Any help would be greatlyappreciated thanks!! After 0.6523g of CoCl2*6H2O is heated,the residue has a mass of 0.4323g. Calculate the % H2Oin the hydrate. What is the actual number of moles of water performula unit CoCl2? I don't even know where to start! Any help would be greatlyappreciated thanks!!

Explanation / Answer

Given data Mass of hydrated salt = 0.6523g Mass of anhydrous salt = 0.4323g Mass of water = 0.6523g - 0.4323 g                          =0.22 g Mass percentage of water in compound =(0.22g/0.6523g)*100                                                              = 33.7 % Moles of anhydrous salt = 0.4323g/ 129.84 g/mol                                       = 0.00333 mol Moles of water = 0.22 g/18.0 g/mol                          = 0.01222 mol ratio of moles of salt to water = 0.00333 mol : 0.0122mol                                               = 1 : 3.67 The actual number of moles of waterperformula unit CoCl2 = 3.67 moles Moles of anhydrous salt = 0.4323g/ 129.84 g/mol                                       = 0.00333 mol Moles of water = 0.22 g/18.0 g/mol                          = 0.01222 mol ratio of moles of salt to water = 0.00333 mol : 0.0122mol                                               = 1 : 3.67 The actual number of moles of waterperformula unit CoCl2 = 3.67 moles

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