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QJEEAdv18Cha53a: JEE Advanced 2018 Chemistry - Class 12 JEEAdv18Cha53a Solution: The ammonia prepared by treating ammonium sulphate with calcium …

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl2.6H2ONiCl{}_2.6{H_2}O to form a stable coordination compound. Assume that both the reactions are 100%100\% complete. If 15841584 gg of ammonium sulphate and 952 g952\,g of NiCl2.6H2ONiC{l_2}.6{H_2}O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is _____________.

(Atomic weights in gg mol−1:mo{l^{ - 1}}: H=1,N=14,O=16,H = 1,N = 14,O = 16, S=32,Cl=35.5,S = 32,Cl = 35.5, Ca=40,Ca = 40, Ni=59Ni = 59

Step-by-Step Verified Solution

JEE Advanced 2018 Paper 1 Offline Chemistry - Coordination Compounds Question 42 English Explanation 1

JEE Advanced 2018 Paper 1 Offline Chemistry - Coordination Compounds Question 42 English Explanation 2

From the above reactions we can see that

132 g of (NH4)2SO4{(N{H_4})_2}S{O_4} will produce 172 g of gypsum

1584 g of (NH4)2SO4{(N{H_4})_2}S{O_4} will produce 172132×1584=2064{{172} \over {132}} \times 1584 = 2064 g of gypsum

Therefore, number of moles of gypsum produced 2064172=12{{2064} \over {172}} = 12 mol

238 g of NiCl4 . 6H2ONiC{l_4}\,.\,6{H_2}O will produce 314 g of Ni-complex

952 g of NiCl4 . 6H2ONiC{l_4}\,.\,6{H_2}O will produce 314238×952=1256{{314} \over {238}} \times 952 = 1256 g of Ni-complex

Therefore, number of moles of Ni-complex produced 1256314=4{{1256} \over {314}} = 4 mol

So, total mass of products is

12 ×\times 172 (gypsum) + 4 ×\times 232 ([Ni(NH3)6]Cl2) = 2992 g

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