strontium chloride can be obtained as the anhydrous salt, SrCl2, and as the hydrates SrCl2.2H20, and SrCl2.6H20. (ArSr = 87.5). 0.323 g of a sample of strontium chloride is dissolved in water and reacted with sliver nitrate solution. The mass of sliver chloride precipitates is 0.476 g. What is the formula of the strontium chloride? ( calulate the amount of the sliver chloride. Since 1 mol of sliver chloride contains 1 mol of chlorine as Cl-, from this you can calulate the mass of the chlorine in the 0.323 g of the strontium chloride.
0.323 g SrCl2.xH2O - g SrCl2 = g H2O.
How much SrCl2 is in the sample.
0.476 g AgCl x (1 mol SrCl2)/2 mol AgCl = ?? g SrCl2.
0.323 - ?? g SrCl2 = xx g H2O
mols SrCl2 = ??g SrCl2/molar mass SrCl2
mols H2O = xx g H2O/molar mass H2O
Find mols H2O per 1 mol SrCl2 and that will be x in the formula SrCl2*xH2O.
Post your work if you get stuck.
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