Chemistry · Coordination Compounds

JEE Advanced 2018 — Paper 1 — Question 20

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by

NiCl2⋅6H2O\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} to form a stable coordination compound. Assume that both the reactions are 100%100 \% complete.

If 1584 g of ammonium sulphate and 952 g of NiCl2⋅6H2O\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{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 gmol−1:H=1, N=14,O=16, S=32,Cl=35.5,Ca=40,Ni=59\mathrm{g} \mathrm{mol}^{-1}: \mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \mathrm{Cl}=35.5, \mathrm{Ca}=40, \mathrm{Ni}=59 )

Answer: 2992

Numerical answer — enter this value.

Step-by-step solution

Weight of gypsum formed =12×172=2064 g=12 \times 172=2064 \mathrm{~g}

NiCl2⋅6H2O+6NH3⟶[Ni(NH3)6]Cl2\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{NH}_{3} \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}

moles 952238=424246=4\frac{952}{238}=4 \quad 24 \quad \frac{24}{6}=4

Mass of [Ni(NH3)6]Cl2\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2} formed =4×232=928 g=4 \times 232=928 \mathrm{~g}

Total weight =2064+928=2992 g=2064+928=2992 \mathrm{~g}.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Chemistry
Chapter
Coordination Compounds
Topic
Properties and importance of Coordination Complexes
The ammonia prepared by treating ammonium sulphate with calcium… | JEE Advanced 2018 PYQ with Solution · DhiX AI