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    See Answer
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