Determine the theoretical yield of H2S (in moles) if 30 mol Al2S3 and 30 mol H2O are reacted according to the following balanced reaction.
Al2S3(s) + 6 H2O(I) → 2 Al(OH)3(s) + 3 H2S(g)
From the balanced equation, we can see that every 1 mol of Al2S3 reacts with 6 mol of H2O to produce 3 mol of H2S.
Since we have 30 mol of Al2S3 and 30 mol of H2O, we can determine the limiting reactant by comparing the moles of each reactant to the stoichiometry of the reaction.
For Al2S3: 30 mol Al2S3 * (3 mol H2S / 1 mol Al2S3) = 90 mol H2S
For H2O: 30 mol H2O * (3 mol H2S / 6 mol H2O) = 15 mol H2S
Since H2O is the limiting reactant, we can determine the theoretical yield of H2S:
15 mol H2S
Therefore, the theoretical yield of H2S is 15 moles.