Subject: Chemistry — Coordination Compounds
Question: The ammonia prepared by treating ammonium sulphate with
calcium hydroxide is completely used by $NiCl_{2}.6H_{2}O$ to form a
stable coordination compound. Assume that both the reactions are 100%
complete. If $1584\ g$ of ammonium sulphate and $952\ g$ of
$NiCl_{2}.6H_{2}O$ are used in the preparation, the combined weight (in
grams) of gypsum and the nickel-ammonia coordination compound thus
produced is \_\_\_\_.
(Atomic weights in $g\ mol^{- 1}$:
$H = 1,\ N = 14,\ O = 16,\ S = 32,\ Cl = 35.5,\ Ca = 40,\ Ni = 59$)
Answer: 2992
Solution: [IMAGE] Number of Moles of
$\left( NH_{4} \right)_{2}SO_{4} = \frac{1584}{132} = 12\ moles$
Moles of $NH_{3}$ released = 24 moles
Moles of $NiCl_{2}.6H_{2}O = \frac{952}{238} = 4\ moles$
Number of moles of Gypsum $\left( CaSO_{4}.2H_{2}O \right)$ formed =
12 moles
Mass of Gypsum formed = 12 × 172 = 2064
Number of moles of complex formed
$\left\lbrack Ni\left( NH_{3} \right)_{6} \right\rbrack Cl_{2} = \frac{24}{6} = 4\ moles$
Mass of complex formed $= \ 4\ \times \ 232\ = \ 928\ g$
Total Mass $= \ 2064\ + \ 928\ = \ 2992\ g$