JEE Challenger
More from Some Basic Concepts of Chemistry
ChemHardAdv 2018 • Paper 1

Combined Mass of Gypsum and Nickel Ammonia Complex Produced

Ammonia generated by reacting ammonium sulphate with calcium hydroxide is quantitatively utilized by NiCl2⋅6H2O\text{NiCl}_2\cdot 6\text{H}_2\text{O} to yield a stable coordination compound. Assuming both consecutive reactions proceed with 100%100\% yield, if 1584 g1584\text{ g} of ammonium sulphate and 952 g952\text{ g} of NiCl2⋅6H2O\text{NiCl}_2\cdot 6\text{H}_2\text{O} are supplied for this preparation, calculate the combined weight (in grams) of gypsum and the nickel-ammonia coordination complex formed.

(Given atomic masses in g mol−1\text{g mol}^{-1}: H=1\text{H} = 1, N=14\text{N} = 14, O=16\text{O} = 16, S=32\text{S} = 32, Cl=35.5\text{Cl} = 35.5, Ca=40\text{Ca} = 40, Ni=59\text{Ni} = 59)

Official Numerical Answer2992

Topics & Concepts

Some Basic Concepts of ChemistryMole ConceptStoichiometryCoordination Compounds

Step-by-Step Solution

To find the combined weight of gypsum and the nickel-ammonia coordination compound, we analyze the two chemical reactions step by step:

Step 1: Reaction of ammonium sulphate with calcium hydroxide

The reaction between ammonium sulphate and calcium hydroxide produces ammonia, water, and calcium sulphate: (NH4)2SO4+Ca(OH)2→CaSO4+2NH3+2H2O(\text{NH}_4)_2\text{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 + 2\text{NH}_3 + 2\text{H}_2\text{O}

In the presence of water, calcium sulphate crystallizes as gypsum (CaSO4⋅2H2O\text{CaSO}_4\cdot 2\text{H}_2\text{O}): (NH4)2SO4+Ca(OH)2→CaSO4⋅2H2O+2NH3(\text{NH}_4)_2\text{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4\cdot 2\text{H}_2\text{O} + 2\text{NH}_3

Molar masses:

  • Molar mass of (NH4)2SO4(\text{NH}_4)_2\text{SO}_4: Molar mass=2×(14+4×1)+32+4×16=2×18+32+64=132 g mol−1\text{Molar mass} = 2 \times (14 + 4 \times 1) + 32 + 4 \times 16 = 2 \times 18 + 32 + 64 = 132\text{ g mol}^{-1}
  • Number of moles of (NH4)2SO4(\text{NH}_4)_2\text{SO}_4 supplied: n(NH4)2SO4=1584 g132 g mol−1=12 moln_{(\text{NH}_4)_2\text{SO}_4} = \frac{1584\text{ g}}{132\text{ g mol}^{-1}} = 12\text{ mol}

From the stoichiometry of the reaction:

  • Moles of NH3\text{NH}_3 produced: nNH3=2×n(NH4)2SO4=2×12=24 moln_{\text{NH}_3} = 2 \times n_{(\text{NH}_4)_2\text{SO}_4} = 2 \times 12 = 24\text{ mol}
  • Moles of gypsum (CaSO4⋅2H2O\text{CaSO}_4\cdot 2\text{H}_2\text{O}) formed: ngypsum=n(NH4)2SO4=12 moln_{\text{gypsum}} = n_{(\text{NH}_4)_2\text{SO}_4} = 12\text{ mol}

Mass of gypsum formed:

  • Molar mass of CaSO4⋅2H2O\text{CaSO}_4\cdot 2\text{H}_2\text{O}: Molar mass=40+32+(4×16)+2×(2×1+16)=136+36=172 g mol−1\text{Molar mass} = 40 + 32 + (4 \times 16) + 2 \times (2 \times 1 + 16) = 136 + 36 = 172\text{ g mol}^{-1}
  • Mass of gypsum: mgypsum=12 mol×172 g mol−1=2064 gm_{\text{gypsum}} = 12\text{ mol} \times 172\text{ g mol}^{-1} = 2064\text{ g}

Step 2: Reaction of nickel chloride with ammonia

The stable coordination compound formed by Ni2+\text{Ni}^{2+} in the presence of excess ammonia is the octahedral complex hexaamminenickel(II) chloride, [Ni(NH3)6]Cl2[\text{Ni(NH}_3)_6]\text{Cl}_2: NiCl2⋅6H2O+6NH3→[Ni(NH3)6]Cl2+6H2O\text{NiCl}_2\cdot 6\text{H}_2\text{O} + 6\text{NH}_3 \rightarrow [\text{Ni(NH}_3)_6]\text{Cl}_2 + 6\text{H}_2\text{O}

Molar masses and moles:

  • Molar mass of NiCl2⋅6H2O\text{NiCl}_2\cdot 6\text{H}_2\text{O}: Molar mass=59+2×35.5+6×18=59+71+108=238 g mol−1\text{Molar mass} = 59 + 2 \times 35.5 + 6 \times 18 = 59 + 71 + 108 = 238\text{ g mol}^{-1}
  • Number of moles of NiCl2⋅6H2O\text{NiCl}_2\cdot 6\text{H}_2\text{O} supplied: nNiCl2⋅6H2O=952 g238 g mol−1=4 moln_{\text{NiCl}_2\cdot 6\text{H}_2\text{O}} = \frac{952\text{ g}}{238\text{ g mol}^{-1}} = 4\text{ mol}

To react completely with 4 mol4\text{ mol} of NiCl2⋅6H2O\text{NiCl}_2\cdot 6\text{H}_2\text{O}, the required amount of NH3\text{NH}_3 is: nNH3 required=6×4=24 moln_{\text{NH}_3\text{ required}} = 6 \times 4 = 24\text{ mol} Since exactly 24 mol24\text{ mol} of NH3\text{NH}_3 is generated from Step 1, both reactants are consumed quantitatively in stoichiometric amounts.

  • Moles of [Ni(NH3)6]Cl2[\text{Ni(NH}_3)_6]\text{Cl}_2 produced: ncomplex=4 moln_{\text{complex}} = 4\text{ mol}

Mass of coordination compound formed:

  • Molar mass of [Ni(NH3)6]Cl2[\text{Ni(NH}_3)_6]\text{Cl}_2: Molar mass=59+6×17+2×35.5=59+102+71=232 g mol−1\text{Molar mass} = 59 + 6 \times 17 + 2 \times 35.5 = 59 + 102 + 71 = 232\text{ g mol}^{-1}
  • Mass of the complex: mcomplex=4 mol×232 g mol−1=928 gm_{\text{complex}} = 4\text{ mol} \times 232\text{ g mol}^{-1} = 928\text{ g}

Step 3: Combined Weight

Combined weight=mgypsum+mcomplex=2064 g+928 g=2992 g\text{Combined weight} = m_{\text{gypsum}} + m_{\text{complex}} = 2064\text{ g} + 928\text{ g} = 2992\text{ g}

PreviousNext