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posted by ken .
Al2(s04)3 solution of a molal concentration is present in 1 litre solution of 2.684g/cc. How many moles of bas04 would be precipitated on adding bacl2 in excess?
the answer 6M
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Al2(SO4)3 +3BaCl2 >> 3BaSO4 (s) + 2AlCl3
first, figure out the number of moles per liter of aluminum sulfate there are in 2.684Kg.
You will get three times that number of moles of BaSO4 (look at the balanced equation).