Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide.
Write the balanced chemical equation for this reaction.
How many grams of aluminum hydroxide are obtained from 16.2 of aluminum sulfide?
The balanced equation is: Al2S3(s) + 6H2O(l) %u2192 2Al(OH)3(s) + 3H2S(g).
It shows that each molecule of aluminum sulfide produces two molecules of aluminum hydroxide, so each mole of Al2S3 produces 2 moles of Al(OH)3.
The molecular weight of Al2S3 is 150 and that of Al(OH)3 is 78. If we start with 16.2 grams of aluminum sulfide we can set up the proportion:16.2 / 150 = x / 156,
where x is the number of grams of Al(OH)3 produced.
We use 156, which is twice 78, since two moles of Al(OH)3 are produced for each mole of Al2S3 that reacts. So we get x = (16.2)(156) / 150 = 16.8 gm. We obtain 16.8 grams of aluminum hydroxide to the nearest decigram.