Question

You are given 3.1569 g of a mixture of nickel(II) chloride hexahydrate and nickel(II) sulfate hexahydrate. You dissolve the mixture in 400 mL of deionised water, and add an excess of silver nitrate. You filter the precipitate off and dry it in an oven. The mass of the dry precipitate is 3.4088 g. How many grams of nickel(II) sulfate hexahydrate are in the mixture?

Answer 1

Nickel sulfate hexahydrate molar mass 262.85

Nickel chloride hexahydrate molar mass 237.69

Moles =$\frac{3.4088}{237.69}=0.014moles$

Mass =$0.014×262.85=3.6799grams$