A certain dry mixture contains only three compounds- silver nitrate, magnesium c

Question

A certain dry mixture contains only three compounds- silver nitrate, magnesium chloride, and barium nitrate. When 1.253 g of the mixture was added to excess water, 0.114 g of dry precipitate was recovered by filteration. The solution which passed through the filter was collected in a beaker and excess sodium sulfate solution was added to this beaker. 0.432 g of fry precipitate was collected from this second solution by filtration. Calculate the percent of magnesium chloride in the original mixture.

Explanation / Answer

The 1.253g is added to water. The precipitate left will be silver chloride. you have to assume that there is enough chloride ion to preciptate all of the silver. The filtrate contains only Mg and Ba cations now. On adding the Na2SO4 (not nitrate) the precipitate will be BaSO4 which amounts to 0.432 g Now you can work backwards and establish how much MgCl2 was present in the original mixture. The substances were AgNO3 and Ba(NO3)2 Mass of AgNO3 = 0.000795 x 169.87 = 0.1352 g Mass of Ba(NO3)2 = 0.00185 x 261.37 = 0.4835 g Mass of MgCl2 must therefore be 1.253 - 0.1352 - 0.4835 = 0.6343 g % of MgCl2 = 0.6343/1.253 x 100 = 50.6%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.