JEE Advanced 2018 · Paper 1 · Chemistry Q8

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ChemistryNumerical+3Question 26 of 54

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $\ce{NiCl2.6H2O}$ to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of $\ce{NiCl2.6H2O}$ 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)

Your answer

Answer and worked solution

Official answer: 2989

Ammonia and gypsum.
$\ce{(NH4)2SO4 + Ca(OH)2 + 2H2O -> CaSO4.2H2O + 2NH3}$
$M[\ce{(NH4)2SO4}]=132$ , so 1584 g is 12 mol. This gives 24 mol $\ce{NH3}$ and 12 mol gypsum ( $M=172$ ):
$12\times172=2064\text{ g}.$

Complex. $M[\ce{NiCl2.6H2O}]=59+71+108=238$ , so 952 g is 4 mol.
$\ce{NiCl2.6H2O + 6NH3 -> [Ni(NH3)6]Cl2 + 6H2O}$
4 mol needs exactly 24 mol $\ce{NH3}$ . $M[\ce{[Ni(NH3)6]Cl2}]=59+102+71=232$ , so
$4\times232=928\text{ g}.$

Total. $2064+928=2992$ g.

Answer: 2992 (key: 2989 to 2995)

Topics:Stoichiometry Nomenclature

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