1) According to the reaction,
1 mole Al2S3 require 6 mole H2O
150 g Al2S3 require 108 g H2O
30 g Al2S3 require 108×30/150 g H2O
= 21.6 g H2O
Since we have only 20 g H2O thus, H2O is limiting reagent.
According to the reaction,
6 mole H2O give 3 mole H2S
108 g H2O give 3×34 = 102 g H2S
20 g H2O give 102×20/108 g H2S
= 18.89 g H2S
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