Discussion Forum

1)

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 952 g of NiCl₂ .6H₂O are used in the preparation, the combined weight (in grams)of gypsum and the nickel -ammonia coordination compound thus produced is  ......

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

Answer:

Option A

Explanation:

Balanced equations of reactions used in the problem are follows,

(i)  $(NH_{4})_{2}SO_{4}+Ca(OH)_{2}\rightarrow CaSO_{4}.2H_{2}O+2NH_{3}$

          1 mol (132g)                                        1 mol (172 g)           2 mol (2×17=34g)

(ii)    $Nicl_{2}.6H_{2}O+6NH_{3}\rightarrow [Ni(NH_{3})_{6}]Cl_{2}+6H_{2}O$

              1mol (238g)       6mol (102g)                  1 mol (232 g)

  Now, in Eq .(i)

     If, 1584 g of ammonium sulphate is used.

i.e, 1584 g of (NH₄)₂ SO₄ = $\frac{1584}{132}=12 mol$

So, accordingto the Eq .(i)  given above 12 moles of (NH₄)₂ SO₄ produces 

(a) 12  moles of gypsum

 (b) 24 moles of ammonia

         Here , 12 moles of gypsum= 12×172=2064 g

and   24 moles of NH₃=24 × 17=408 g

 Further , as given in question,

      24 moles of NH₃ produced in reaction (i) is completly utilised by 952 g or 4 moles of NiCl_2. 6H₂O to produce 4 moles of [Ni(NH₃)₆ ]Cl₂.

  So 4 moles of [Ni (NH₃)₆ ]Cl₂ =4× 232 =928 gms

Hence, total mass of gypsum and nickel ammonia coordination compound [Ni(NH₃)]₆Cl₂

                                         =2064+928=2992