Ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used `NiCl_2.6H_2O` to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of `NiCl_2.6H_2O` are used in the preparation, the combined weight (in grams) of gypsum and the nickel - ammonia coordination compound thus produced is _____.
(Atomic weights in g `mol^(-1)` : H=1,N=14,O=16 , S=32, Cl=35.5 , Ca=40 , Ni=59)

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the reactions
The first reaction involves the preparation of ammonia from ammonium sulfate and calcium hydroxide:
\[ \text{(NH}_4\text{)}_2\text{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 \cdot 2\text{H}_2\text{O} + 2\text{NH}_3 \]

The second reaction involves the formation of a nickel-ammonia coordination compound:
\[ \text{NiCl}_2 \cdot 6\text{H}_2\text{O} + 6\text{NH}_3 \rightarrow \text{Ni(NH}_3\text{)}_6\text{Cl}_2 + 6\text{H}_2\text{O} \]

### Step 2: Calculate the molecular weights
1. **Ammonium sulfate \((\text{NH}_4)_2\text{SO}_4\)**:
- \(2 \times 14 + 8 + 32 + 16 \times 4 = 132 \, \text{g/mol}\)

2. **Gypsum \((\text{CaSO}_4 \cdot 2\text{H}_2\text{O})\)**:
- \(40 + 32 + 16 \times 4 + 2 \times 18 = 172 \, \text{g/mol}\)

3. **Nickel(II) chloride hexahydrate \((\text{NiCl}_2 \cdot 6\text{H}_2\text{O})\)**:
- \(59 + 2 \times 35.5 + 6 \times 18 = 232 \, \text{g/mol}\)

### Step 3: Calculate moles of reactants
1. **Moles of ammonium sulfate**:
\[ \text{Moles of } (\text{NH}_4)_2\text{SO}_4 = \frac{1584 \, \text{g}}{132 \, \text{g/mol}} = 12 \, \text{moles} \]

2. **Moles of nickel(II) chloride hexahydrate**:
\[ \text{Moles of } \text{NiCl}_2 \cdot 6\text{H}_2\text{O} = \frac{952 \, \text{g}}{232 \, \text{g/mol}} = 4 \, \text{moles} \]

### Step 4: Calculate the weight of gypsum produced
According to the stoichiometry of the first reaction, 1 mole of ammonium sulfate produces 1 mole of gypsum. Therefore, 12 moles of ammonium sulfate will produce 12 moles of gypsum.

\[ \text{Weight of gypsum} = 12 \, \text{moles} \times 172 \, \text{g/mol} = 2064 \, \text{g} \]

### Step 5: Calculate the weight of the nickel-ammonia coordination compound produced
From the second reaction, 1 mole of nickel(II) chloride hexahydrate produces 1 mole of the nickel-ammonia coordination compound. Therefore, 4 moles of nickel(II) chloride will produce 4 moles of the coordination compound.

\[ \text{Weight of Ni(NH}_3\text{)}_6\text{Cl}_2 = 4 \, \text{moles} \times 232 \, \text{g/mol} = 928 \, \text{g} \]

### Step 6: Calculate the combined weight of gypsum and the nickel-ammonia coordination compound
\[ \text{Combined weight} = \text{Weight of gypsum} + \text{Weight of Ni(NH}_3\text{)}_6\text{Cl}_2 \]
\[ \text{Combined weight} = 2064 \, \text{g} + 928 \, \text{g} = 2992 \, \text{g} \]

### Final Answer
The combined weight of gypsum and the nickel-ammonia coordination compound produced is **2992 grams**.

---