Question

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 ($\left[Ni\right(N{H}_3{)}_6]C{l}_2$ ). 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 ($CaS{O}_4.2H_{2}O$) and the nickel-ammonia coordination compound thus produced is ____.

(The reactions involved are:
$Ca(OH{)}_2+(N{H}_4{)}_{2}S{O}_4\to 2N{H}_3+CaS{O}_4.2H_{2}O$
$NiC{l}_2.6H_{2}O+6N{H}_3\to \left[Ni\right(N{H}_3{)}_6]C{l}_2+6H_{2}O$

Atomic weights in $gmo{l}^{-1}$ : $H=1,N=14,O=16,S=32,Cl=35.5,Ca=40,Ni=59)$

• A. 2929 g
• B. 2992 g
• C. 2998 g
• D. 2982 g

Answer 1

The correct option is B 2992 g

The reaction for the ammonia formation is $Ca(OH{)}_2+(N{H}_4{)}_{2}S{O}_4\to 2N{H}_3+CaS{O}_4.2H_{2}O$
$moles:12\frac{1584}{132}=122412$

$\therefore$ Weight of gypsum $\left(CaS{O}_4.2H_{2}O\right)$ formed = $12\times 172=2064g$

Again the reaction corresponding to the complex formation is:


$NiC{l}_2.6H_{2}O+6N{H}_3\to \left[Ni\right(N{H}_3{)}_6]C{l}_2+6H_{2}O$
$moles:\frac{952}{238}=424\frac{24}{6}=4$

$\therefore$Mass of $\left[Ni\right(N{H}_3{)}_6]C{l}_2$ formed = $4\times 232=928g$
Total Weight = $2064+928=2992g$