Consider a 1.50g mixture of magnesium nitrate (Mg(NO3)2) and magnesium chloride (MgCl2). After dissolving this mixture in water, 0.500 M silver nitrate (AgNO3) is added dropwise until precipitate formation is complete. The mass of the white precipitate formed is 0.641g.
a) Calculate the mass percent of the magnesium chloride in the mixture.
b) Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.
2007-09-13 09:32:50 · 1 answers · asked by tay 3 in Science & Mathematics ➔ Chemistry
2AgNO3 + MgCl2 ===> 2AgCl + Mg(NO3)
Atomic weights: Ag=108 Cl=35.5 AgCl=143.5 Mg=24 MgCl2=95 Let the mixture be called M.
0.641gAgCl x 1molAgCl/143.5gAgCl x 1molMgCl2/2molAgCl x95gMgCl2/1molMgCl2 = 0.212g MgCl2
0.212gMgCl2/1.50gM x 100% = 14.1 mass-% MgCl2
0.641gAgCl x 1molAgCl/143.5gAgCl x 1molAgNO3/1molAgCl x 1000mLAgNO3/0.500molAgNO3 = 8.93
2007-09-13 09:59:37 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋