The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl2.6H2O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952g of NiCl2.6H2O are used in the preparation, the combined weight (in kg) of gypsum and the nickel-ammonia coordination compound thus produced is___. [JEE 2018] ( 4 )2 4+ ( )2 ® 4 2 + 3 NH SO Ca OH CaSO .2H O 2NH 2 × 2 + 3 ®éë ( 3 )6 ùû 2 + 2 NiCl 6H O 6NH Ni NH Cl 6H O (Atomic weights in g mol–1: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)

Key Concepts Needed

1. Molar Mass (Gram-Molecular Weight)
   The mass of one mole of a substance. You find it by adding the atomic masses of all atoms in the formula.
2. Stoichiometry
   Balanced chemical equations tell us the exact mole ratio in which reactants combine and products form.
3. Limiting Reagent
   The reactant that finishes first stops the reaction. Everything else is decided by this reagent.
4. Hydrates & Coordination Compounds
   • Gypsum: $CaSO_4 \cdot 2H_2O$ (has water trapped in its crystal)
   • Coordination complex: $[Ni(NH_3)_6]Cl_2 \cdot 6H_2O$

Logical Chain to Tackle the Problem

1. Write the balanced equations supplied in the question.
2. Compute molar masses of all species.
3. Convert given grams to moles.
4. Use mole ratios to see how much ammonia is produced and how much is required.
5. Identify whether any reactant is left over (limiting reagent concept).
6. Calculate moles of final products based on the limiting reagent.
7. Convert moles of products back to grams, add them, and convert to kilograms for the final answer.

Imagine This Like Making Sandwiches and Lemonade

1. Bread + Butter ➜ Sandwich + Free Chocolate Here, ammonium sulphate $(NH_4)_2SO_4$ mixes with calcium hydroxide $Ca(OH)_2$ to give gypsum (something solid like a sandwich) plus free ammonia gas (tasty chocolates).
2. Chocolate + Lemon Syrup ➜ Chocolate-Lemon Candy
   Next, the free ammonia (chocolate) reacts with a nickel salt (lemon syrup) to make a fancy nickel candy (a shiny blue crystal).

We count how many slices of bread, how much butter, chocolates, and syrup we have so that nothing is wasted—just like making sure every kid in the party gets exactly one sandwich and one candy. Finally, we weigh all the sandwiches and candies together to know the total goodies we produced.

Step-by-Step Calculation

1. Molar Masses
   $M_{(NH_4)_2SO_4}=\left[2(14+4) + 32 + 4\times16\right]=132\,g\,mol^{-1}$
   $M_{NiCl_2\cdot6H_2O}=59 + 2\times35.5 + 6\times18 = 238\,g\,mol^{-1}$
   $M_{CaSO_4\cdot2H_2O}=40 + 32 + 4\times16 + 2\times18 = 172\,g\,mol^{-1}$
   $M_{[Ni(NH_3)_6]Cl_2\cdot6H_2O}=59 + 6(14+3) + 2\times35.5 + 6\times18 = 340\,g\,mol^{-1}$

2. Convert Given Mass to Moles
   $n_{(NH_4)_2SO_4}=\frac{1584}{132}=12\,mol$
   $n_{NiCl_2\cdot6H_2O}=\frac{952}{238}=4\,mol$

3. Moles of $NH_3$ Produced
   From the first reaction, $1\,mol (NH_4)_2SO_4 \rightarrow 2\,mol\,NH_3$
   $n_{NH_3}=12\times2=24\,mol$

4. Moles of $NH_3$ Required
   From the second reaction, $1\,mol\,NiCl_2\cdot6H_2O\;\text{needs}\;6\,mol\,NH_3$
   $n_{NH_3\,\text{needed}}=4\times6=24\,mol$ Both match → No excess; all reactions 100% complete.

5. Moles & Mass of Products
   • Gypsum: $n_{CaSO_4\cdot2H_2O}=12\,mol$
   $m=12\times172=2064\,g$
   • Nickel complex: $n_{[Ni(NH_3)_6]Cl_2\cdot6H_2O}=4\,mol$
   $m=4\times340=1360\,g$

6. Total Mass of Products
   $m_{total}=2064\,+\,1360=3424\,g=3.424\,kg$

Answer: 3.424 kg (≈3.42 kg)