Calculate the number of milliliters of 0.469 M
Ba(OH)2 required to precipitate all of
the Al3+ ions...
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Chemistry
Calculate the number of milliliters of 0.469 MBa(OH)2 required to precipitate all ofthe Al3+ ions in 179mL of 0.565 M AlCl3solution as Al(OH)3. The equation forthe reaction is:
2AlCl3(aq) +3Ba(OH)2(aq)Al(OH)3(s) +3BaCl2(aq)
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4.4 Ratings (600 Votes)
From the eqation below3 mols of BaOH2 is required to precipitate 2 mols ofAlCl3We are given with 179mL of 0565 M AlCl3
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