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Al2S3 + 2 Na3PO4 = 2 AlPO4 + 3 Na2S

moles Al2S3 = 0.25 M x 1.00 L = 0.25

moles AlPO4 = 0.25 x 2 = 0.50

mass AlPO4 = 0.50 mol x 121.95 g/mol=61.0 g

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Q: Sodium phosphate in excess is poured into 1.00 L of a solution of aluminium sulphide at 0.25 M Calculate the mass of the precipitate formed?
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