The user wants me to solve a stoichiometry problem involving two sequential chemical reactions.

Step 1: Write down the balanced chemical reactions and identify the products.

Reaction 1: Preparation of ammonia Ammonium sulphate reacts with calcium hydroxide to produce ammonia, calcium sulphate, and water. $(NH_4)_2SO_4 + Ca(OH)_2 \rightarrow CaSO_4 + 2NH_3 + 2H_2O$ The problem asks for the weight of gypsum produced. Gypsum is the common name for calcium sulphate dihydrate, $CaSO_4 \cdot 2H_2O$. We need to check if enough water is produced in the reaction to form the dihydrate. According to the stoichiometry, 1 mole of $(NH_4)_2SO_4$ produces 1 mole of $CaSO_4$ and 2 moles of $H_2O$. This is the exact ratio required for the formation of $CaSO_4 \cdot 2H_2O$. Therefore, the product is indeed gypsum, and the overall reaction is: $(NH_4)_2SO_4 + Ca(OH)_2 \rightarrow CaSO_4 \cdot 2H_2O + 2NH_3$

Reaction 2: Formation of the coordination compound The ammonia produced is used by $NiCl_2 \cdot 6H_2O$ to form a stable coordination compound. Nickel(II) is a $d^8$ ion and typically forms a hexacoordinate octahedral complex with a strong field ligand like ammonia. The compound is $[Ni(NH_3)_6]Cl_2$. The ammonia ligands replace the water ligands. $NiCl_2 \cdot 6H_2O + 6NH_3 \rightarrow [Ni(NH_3)_6]Cl_2 + 6H_2O$

Step 2: Calculate the molar masses of the relevant compounds.

Using the given atomic weights ($H=1, N=14, O=16, S=32, Cl=35.5, Ca=40, Ni=59$):

• Molar mass of ammonium sulphate, $(NH_4)_2SO_4 = 2(14+4) + 32 + 4(16) = 36 + 32 + 64 = 132 \, g/mol$.
• Molar mass of nickel chloride hexahydrate, $NiCl_2 \cdot 6H_2O = 59 + 2(35.5) + 6(18) = 59 + 71 + 108 = 238 \, g/mol$.
• Molar mass of gypsum, $CaSO_4 \cdot 2H_2O = 40 + 32 + 4(16) + 2(18) = 136 + 36 = 172 \, g/mol$.
• Molar mass of the coordination compound, $[Ni(NH_3)_6]Cl_2 = 59 + 6(14+3) + 2(35.5) = 59 + 6(17) + 71 = 59 + 102 + 71 = 232 \, g/mol$.

Step 3: Calculate the moles of reactants.

• Moles of $(NH_4)_2SO_4 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1584 \, g}{132 \, g/mol} = 12 \, mol$.
• Moles of $NiCl_2 \cdot 6H_2O = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{952 \, g}{238 \, g/mol} = 4 \, mol$.

Step 4: Determine the amount of products from the first reaction.

The reaction is $(NH_4)_2SO_4 + Ca(OH)_2 \rightarrow CaSO_4 \cdot 2H_2O + 2NH_3$. The problem states the reaction is $100\%$ complete. We assume $Ca(OH)_2$ is in excess.

• From the stoichiometry, 1 mole of $(NH_4)_2SO_4$ produces 1 mole of $CaSO_4 \cdot 2H_2O$ and 2 moles of $NH_3$.
• Therefore, $12 \, mol$ of $(NH_4)_2SO_4$ will produce:
  • Moles of $CaSO_4 \cdot 2H_2O$ (gypsum) = $12 \, mol$.
  • Moles of $NH_3 = 2 \times 12 = 24 \, mol$.

Step 5: Determine the limiting reactant and amount of product from the second reaction.

The reaction is $NiCl_2 \cdot 6H_2O + 6NH_3 \rightarrow [Ni(NH_3)_6]Cl_2 + 6H_2O$.

• Available moles of reactants: $4 \, mol$ of $NiCl_2 \cdot 6H_2O$ and $24 \, mol$ of $NH_3$.
• The stoichiometric ratio required is $\frac{\text{moles of } NiCl_2 \cdot 6H_2O}{\text{moles of } NH_3} = \frac{1}{6}$.
• The available ratio of moles is $\frac{4}{24} = \frac{1}{6}$.
• Since the available ratio is equal to the stoichiometric ratio, both reactants will be completely consumed (as stated in the problem).
• We can use either reactant to calculate the moles of product. Using $NiCl_2 \cdot 6H_2O$:
  • Moles of $[Ni(NH_3)_6]Cl_2$ formed = Moles of $NiCl_2 \cdot 6H_2O$ used = $4 \, mol$.

Step 6: Calculate the combined mass of the products.

• Mass of gypsum ($CaSO_4 \cdot 2H_2O$) produced: $\text{Mass} = \text{moles} \times \text{Molar Mass} = 12 \, mol \times 172 \, g/mol = 2064 \, g$
• Mass of nickel-ammonia coordination compound ($[Ni(NH_3)_6]Cl_2$) produced: $\text{Mass} = \text{moles} \times \text{Molar Mass} = 4 \, mol \times 232 \, g/mol = 928 \, g$
• Combined weight of the two products: $\text{Total Mass} = 2064 \, g + 928 \, g = 2992 \, g$

The combined weight of gypsum and the nickel-ammonia coordination compound is $2992\,g$.