Some Basic Concepts of Chemistry - Result Question 25
Numerical Value Based Question
25. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $NiCl _2 \cdot 6 H 2 O$ to form a stable coordination compound. Assume that both the reactions are $100 %$ complete. If $1584 g$ of ammonium sulphate and $952 g$ of $NiCl 2 \cdot 6 H 2 O$ are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is
(Atomic weights in $g mol^{-1}: H=1, N=14, O=16, S=32$,
$Cl=35.5, Ca=40, Ni=59$ )
(2018 Adv.)
Read the following questions and answer as per the direction given below:
(a) Statement I is true; Statement II is true; Statement II is the correct explanation of Statement I.
(b) Statement I is true; Statement II is true; Statement II is not the correct explanation of Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
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Answer:
Correct Answer: 25. (2992)
Solution:
Balanced equations of reactions used in the problem are as follows
Now, in Eq. (i)
if, 1584 g of ammonium sulphate is used.
i.e., 1584 g $ (NH_4)_2 SO_4 =\dfrac{1584}{132}=12 \text {mol}$
So, according to the Eq. (i) given above 12 moles of $\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$ produces
(a) 12 moles of gypsum
(b) 24 moles of ammonia
Here, 12 moles of gypsum $=12 \times 172=2064 \mathrm{~g}$
and $\quad 24$ moles of $\mathrm{NH}_3=24 \times 17=408 \mathrm{~g}$
Further, as given in question,
24 moles of $\mathrm{NH}_3$ produced in reaction (i) is completly utilised by 952 g or 4 moles of $\mathrm{NiCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}$ to produce 4 moles of $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_2$.
So, 4 moles of $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_2=4 \times 232=928 \mathrm{gms}$ Hence, total mass of gypsum and nickel ammonia coordination compound $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_2=2064+928=2992$
BETA
