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Question: The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl...

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

(NH₄)₂SO₄ + Ca(OH)₂ → CaSO₄.2H₂O+ 2NH₃ NiCl₂.6H₂O+6NH₃ → [Ni(NH₃)₆]Cl₂+6H₂O

[JEE 2018]

(Atomic weights in g mol⁻¹: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)

Answer

2.992

Explanation

Solution

The problem involves two sequential chemical reactions. We need to calculate the mass of the products formed and then find their combined weight.

Step 1: Write down the balanced chemical equations for both reactions.

  1. Preparation of ammonia and gypsum: (NH₄)₂SO₄ + Ca(OH)₂ → CaSO₄·2H₂O + 2NH₃
  2. Formation of nickel-ammonia coordination compound: NiCl₂·6H₂O + 6NH₃ → [Ni(NH₃)₆]Cl₂ + 6H₂O

Step 2: Calculate the molar masses of the reactants and products involved.

  • Molar mass of (NH₄)₂SO₄ = 2(14) + 8(1) + 32 + 4(16) = 28 + 8 + 32 + 64 = 132 g/mol
  • Molar mass of CaSO₄·2H₂O (Gypsum) = 40 + 32 + 4(16) + 2(18) = 40 + 32 + 64 + 36 = 172 g/mol
  • Molar mass of NiCl₂·6H₂O = 59 + 2(35.5) + 6(18) = 59 + 71 + 108 = 238 g/mol
  • Molar mass of [Ni(NH₃)₆]Cl₂ = 59 + 6(14 + 3*1) + 2(35.5) = 59 + 6(17) + 71 = 59 + 102 + 71 = 232 g/mol

Step 3: Calculate the moles of the given reactants.

  • Moles of (NH₄)₂SO₄ = Given mass / Molar mass = 1584 g / 132 g/mol = 12 mol
  • Moles of NiCl₂·6H₂O = Given mass / Molar mass = 952 g / 238 g/mol = 4 mol

Step 4: Calculate the moles and mass of products from Reaction 1. From the first balanced equation: 1 mol (NH₄)₂SO₄ produces 1 mol CaSO₄·2H₂O and 2 mol NH₃.

  • Since 12 mol of (NH₄)₂SO₄ are used, 12 mol of CaSO₄·2H₂O (gypsum) will be produced.
  • Mass of CaSO₄·2H₂O = Moles × Molar mass = 12 mol × 172 g/mol = 2064 g
  • The NH₃ produced will be 2 × 12 mol = 24 mol. This NH₃ is completely used in the second reaction.

Step 5: Calculate the moles and mass of products from Reaction 2. From the second balanced equation: 1 mol NiCl₂·6H₂O reacts with 6 mol NH₃ to produce 1 mol [Ni(NH₃)₆]Cl₂.

  • We have 4 mol of NiCl₂·6H₂O.
  • The NH₃ required for 4 mol of NiCl₂·6H₂O to react completely is 4 mol × 6 = 24 mol.
  • Since we have exactly 24 mol of NH₃ available from Reaction 1, both reactants are completely consumed.
  • Therefore, 4 mol of [Ni(NH₃)₆]Cl₂ will be produced.
  • Mass of [Ni(NH₃)₆]Cl₂ = Moles × Molar mass = 4 mol × 232 g/mol = 928 g

Step 6: Calculate the combined weight of the products. Combined weight = Mass of CaSO₄·2H₂O + Mass of [Ni(NH₃)₆]Cl₂ Combined weight = 2064 g + 928 g = 2992 g

Step 7: Convert the combined weight to kilograms. Combined weight in kg = 2992 g / 1000 g/kg = 2.992 kg

The final answer is 2.992\boxed{\text{2.992}}.

Explanation of the solution:

  1. Calculate molar masses of all reactants and products: (NH₄)₂SO₄ (132 g/mol), CaSO₄·2H₂O (172 g/mol), NiCl₂·6H₂O (238 g/mol), [Ni(NH₃)₆]Cl₂ (232 g/mol).
  2. Determine moles of initial reactants: (NH₄)₂SO₄ (1584 g / 132 g/mol = 12 mol), NiCl₂·6H₂O (952 g / 238 g/mol = 4 mol).
  3. From the first reaction, 12 mol of (NH₄)₂SO₄ yields 12 mol of CaSO₄·2H₂O and 24 mol of NH₃.
  4. Mass of CaSO₄·2H₂O = 12 mol × 172 g/mol = 2064 g.
  5. From the second reaction, 4 mol of NiCl₂·6H₂O requires 4 × 6 = 24 mol of NH₃. Since 24 mol of NH₃ is available, 4 mol of [Ni(NH₃)₆]Cl₂ is formed.
  6. Mass of [Ni(NH₃)₆]Cl₂ = 4 mol × 232 g/mol = 928 g.
  7. Total combined weight = 2064 g + 928 g = 2992 g.
  8. Convert to kg: 2992 g = 2.992 kg.