To solve this problem, we need to use stoichiometry to determine the limiting reactant and the theoretical yield of the product.
First, we need to determine the limiting reactant by calculating the number of moles of each reactant:
Number of moles of NH3 = mass / molar mass = 25 g / 17.03 g/mol = 1.47 mol
Number of moles of H2S = mass / molar mass = 96 g / 34.08 g/mol = 2.82 mol
According to the balanced equation, the reaction requires 2 moles of NH3 for every mole of H2S. Therefore, NH3 is the limiting reactant since we only have 1.47 moles of NH3, while we need 2.00 moles to react with all of the H2S.
Now, we can use the number of moles of NH3 to calculate the theoretical yield of (NH4)2S:
Number of moles of (NH4)2S = 1.47 mol NH3 x (1 mol (NH4)2S / 2 mol NH3) = 0.735 mol (NH4)2S
Finally, we can use the molar mass of (NH4)2S to convert the number of moles to mass:
Mass of (NH4)2S = number of moles x molar mass = 0.735 mol x 68.15 g/mol = 50.0 g
Therefore, the maximum mass of ammonium sulfide that can be formed is 50.0 g.