Question

10.222 g sample of hydrated barium iodide (Bal) is heated to dry off the water of

Answer 1

Answer:

                     BaI₂.5H₂O

Explanation:

Given Data:

                   Mass of Hydrated BaI₂ = 10.222 g

                   Mass of dried BaI₂ = 9.520 g

                   Mass of Water removed  =  10.222-9.520 = 0.702 g

                   M.Mass of BaI₂ = 391.136 g/mol

                   M.Mass of Water  =  18.02 g/mol

Now,

         Calculate moles of dried BaI₂ as,

Moles  =  Mass / M.Mass

Moles  =  9.520 g / 391.136 g/mol

Moles  =  0.02434 moles

         Calculate moles of Water as,

Moles  =  Mass / M.Mass

Moles  =  0.702 g / 18.02 g/mol

Moles  =  0.0389 moles

Then,    

          Calculate Mole ratio of BaI₂ and water as,

              = 0.02434 moles BaI₂ / 0.0389 moles Water

              =  0.625

Now,

We will convert this mole ratio to a whole number by multiplying it with a nearest integer,

             =  0.625 × 8

             =  5

Hence, this means for every one mole of BaI there are 5 moles of Water.

Result:

                                             BaI₂.5H₂O