Question

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $NiC{l}_2.6H_{2}O$ to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of $NiC{l}_2.6H_{2}O$ 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 $mo{l}^{-1}$ : $H$ = 1, $N$ = 14, $O$ = 16, $S$ = 32, $Cl$ = 35.5, $Ca$ = 40, $Ni$ = 59)

Answer 1

$\begin{array}{c}\\ moles\end{array}\begin{array}{c}Ca(OH{)}_2+\\ 12\end{array}\begin{array}{c}(N{H}_4{)}_{2}S{O}_4\\ \frac{1584}{132}=12\end{array}\to \begin{array}{c}2N{H}_3+\\ 24\end{array}\begin{array}{c}CaS{O}_4.2H_{2}O\\ \left(Gypsum\right)\end{array}$
Weight of gypsum formed $=12\times 172=2064g$
$\begin{array}{c}\\ moles\end{array}\begin{array}{c}NiC{l}_2.6H_{2}O+\\ \frac{952}{238}=4\end{array}\begin{array}{c}6N{H}_3\\ 24\end{array}\to \begin{array}{c}\left[Ni\right(N{H}_3{)}_6]C{l}_2\\ \frac{24}{6}=4\end{array}$
Mass of $\left[Ni\right(N{H}_3{)}_6]C{l}_{2}formed=4\times 232=928g$
Total weight =2064 + 928 = 2992 g.