for the first question NH₃ is the limiting reagent, if you use the stoichiometry of the reaction. 2NH₃ produce 1Mg(OH)2 and there for the value5.4 mols of 2NH₃ will give us half of Mg(OH)2 which is 2.7 mols. while the ratio of the magnesium sulphate and magnesium hydroxide is 1:1 there 4.6 mols of magnesium sulphate will give the higher amount of magnisium hydroxide.
for the second question, in order to find the greatest amount of magnesium hydroxide we will use the reactant that is in excess which is magnesium sulphate. so we know that n = m/M where n is the number of mols, m is the mass and M is the molar mass. we are given 4.6 mols of magnesium sulphate so m = nxM
m = (4.6 mols)x (58.32g/mol) = 268.27 grams of Mg(OH)2 will be produce
for the last question, to find the excess mols after completion we first find the amount og magnesium hydroxide produced using the limiting reagent and determine the mols remaining. using the limiting reagent NH₃ 2.7 mols will react therefor the mass that we get is m = nxM (2.7 mols)x(58.32g/mol) = 157.46 grams will be formed. from this we now know that 2.7 mols of the MgSO4 will also react leaving behind 4,6 - 2,7 = 1.9 mol of MgSO4 as excess mols.