Amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted?

... A few of the molar masses are as follows:

Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol.

Al2S3 (s) + 6 H2O (l) → 2 Al (OH) 3 (s) + 3 H2S (g)

The chemical reaction is given as:

Al2S3 (s) + 6 H2O (l) → 2 Al (OH) 3 (s) + 3 H2S (g)

We are given the amount of the reactants to be used for the reaction. We need to convert these masses to determine the limiting reactant to use for the calculations.

2.00 g Al2S3 (1 mol / 150.17 g) = 0.0134 mol Al2S3

2.00 g H2O (1 mol / 18.02 g) = 0.1110 mol H2O

Therefore, the limiting reactant is Al2S3 since it is what is consumed completely in the reaction. Hope this answers the question. Have a nice day.