JEE Advance - Chemistry (2018 - Paper 1 Offline - No. 7)
The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $$NiCl{}_2.6{H_2}O$$ to form a stable coordination compound. Assume that both the reactions are $$100\% $$ complete. If $$1584$$ $$g$$ of ammonium sulphate and $$952\,g$$ of $$NiC{l_2}.6{H_2}O$$ are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is _____________.
(Atomic weights in $$g$$ $$mo{l^{ - 1}}:$$ $$H = 1,N = 14,O = 16,$$ $$S = 32,Cl = 35.5,$$ $$Ca = 40,$$ $$Ni = 59$$
(Atomic weights in $$g$$ $$mo{l^{ - 1}}:$$ $$H = 1,N = 14,O = 16,$$ $$S = 32,Cl = 35.5,$$ $$Ca = 40,$$ $$Ni = 59$$
Answer
2992
Explanation
From the above reactions we can see that
132 g of $${(N{H_4})_2}S{O_4}$$ will produce 172 g of gypsum1584 g of $${(N{H_4})_2}S{O_4}$$ will produce $${{172} \over {132}} \times 1584 = 2064$$ g of gypsum
Therefore, number of moles of gypsum produced $${{2064} \over {172}} = 12$$ mol
238 g of $$NiC{l_4}\,.\,6{H_2}O$$ will produce 314 g of Ni-complex
952 g of $$NiC{l_4}\,.\,6{H_2}O$$ will produce $${{314} \over {238}} \times 952 = 1256$$ g of Ni-complex
Therefore, number of moles of Ni-complex produced $${{1256} \over {314}} = 4$$ mol
So, total mass of products is
12 $$\times$$ 172 (gypsum) + 4 $$\times$$ 232 ([Ni(NH3)6]Cl2) = 2992 g
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