Balanced equations of reactions used in the problem are as follows (i) (NH4​)2​SO4​+Ca(OH)2​⟶1 mol132 g (ii) NiCl2​⋅6H2​O+1 mol238 g​CaSO4​⋅2H2​O+1 mol172 g6NH3​⟶6 mol102 g​2NH3​−2 mol(2×17)=34 g[Ni(NH3​)6​]Cl2​+6H2​O1 mol232 g​Now, in Eq. (i)if, 1584 g of ammonium sulphate is used.i.e., 1584 g(NH4​)2​SO4​=1321584​=12 molSo, according to the Eq. (i) given above 12 moles of (NH4​)2​SO4​ produces(a) 12 moles of gypsum(b) 24 moles of ammoniaHere, 12 moles of gypsum =12×172=2064 gand 24 moles of NH3​=24×17=408 gFurther, as given in question,24 moles of NH3​ produced in reaction (i) is completely utilised by 952 g or 4 moles of NiCl2​⋅6H2​O to produce 4 moles of [Ni(NH3​)]Cl2​.So, 4 moles of [Ni(NH3​)1​]Cl2​=4×232=928gmsHence, total mass of gypsum and nickel ammonia coordination compound [Ni(NH3​)6​]Cl2​=2064+928=2992