Amount of al2s3 remains when 20.00 g of al2s3 and 2.00 g of h2o are reacted?
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.

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