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What mass of precipitate is produced by the reaction of 20.0mL of 0.210 mol/L sodium sulfide with an excess quantity of aluminum nitrate solution?
The two solutions provided react with each other and the resulting precipitate is separated by filtration and dried. The mass of the dried precipitate is determined. A yellow precipitate resembling aluminum sulfide was formed. mass of filter paper = 0.97g mass of filter paper plus precipitate = 1.17g A few additional drops of the sodium sulfide solution added to the filtrate formed precipitate. So far this is what I have for information to solving this problem, I'm lost from here. Na2S + Al(NO3)3 --> NaNO3 + Al2S3 3Na2S + 2Al(NO3)3 --> 6NaNO3 + Al2S3 20.0mL 0.210mol/L Thanks |
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