Determine the theoretical yield of H25 (in moles) if 48 mol AlS and 48 mol H20 are reacted according to the following balanced reaction.
Al2S3(5) + 6 H20(l) → 2 Al(OH)3(s) + 3 H2S(g)
The balanced chemical equation indicates that 1 mole of Al2S3 reacts with 6 moles of H2O to produce 3 moles of H2S.
Given that 48 mol of H2O is reacted, we must first determine the limiting reactant between Al2S3 and H2O.
The molar ratio between Al2S3 and H2O is 1:6, so the number of moles of Al2S3 required to react with 48 mol of H2O is 48 mol H2O * (1 mol Al2S3 / 6 mol H2O) = 8 mol Al2S3.
Since we are given 48 mol of Al2S3, which is more than the amount required for the reaction, Al2S3 will be in excess and H2O will be the limiting reactant.
From the balanced chemical equation, we know that 6 moles of H2O produce 3 moles of H2S. Therefore, 48 mol H2O will produce 48 mol H2S * (3 mol H2S / 6 mol H2O) = 24 mol H2S.
Therefore, the theoretical yield of H2S in moles is 24 mol.