Some Basic Concepts of Chemistry 2 Question 25
25. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $\mathrm{NiCl}{2} \cdot 6 \mathrm{H}{2} \mathrm{O}$ to form a stable coordination compound. Assume that both the reactions are $100 %$ complete. If $1584 \mathrm{~g}$ of ammonium sulphate and $952 \mathrm{~g}$ of $\mathrm{NiCl}{2} \cdot 6 \mathrm{H}{2} \mathrm{O}$ are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is
(Atomic weights in $\mathrm{g} \mathrm{mol}^{-1}: \mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32$,
$\mathrm{Cl}=35.5, \mathrm{Ca}=40, \mathrm{Ni}=59$ )
(2018 Adv.)
Some Basic Concepts of Chemistry
Assertion and Reason
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.
Show Answer
Solution:
- Balanced equations of reactions used in the problem are as follows
(i) $\left(\mathrm{NH}{4}\right){2} \mathrm{SO}{4}+\mathrm{Ca}(\mathrm{OH}){2} \longrightarrow \mathrm{CaSO}{4} \cdot 2 \mathrm{H}{2} \mathrm{O}+\underset{2 \mathrm{~mol}}{2 \mathrm{NH}_{3}}$ $\begin{array}{lcc}1 \mathrm{~mol} & 1 \mathrm{~mol} & 2 \mathrm{~mol} \ 132 \mathrm{~g} & 172 \mathrm{~g} & (2 \times 17)=34 \mathrm{~g}\end{array}$
Now, in Eq. (i)
if, $1584 \mathrm{~g}$ of ammonium sulphate is used.
i.e., $1584 \mathrm{~g}\left(\mathrm{NH}{4}\right){2} \mathrm{SO}_{4}=\frac{1584}{132}=12 \mathrm{~mol}$
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 $952 \mathrm{~g}$ 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 $\left[\mathrm{Ni}\left(\mathrm{NH}{3}\right){6}\right] \mathrm{Cl}_{2}=2064+928=2992$
