Question

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $NiC{l_2}.6{H_2}O$ to form a stable coordination compound. Assume that both the reactions are $100\%$complete. If 1584 gm of ammonium sulphate and 952 gm of $NiC{l_2}.6{H_2}O$ are used in the preparation, the combined weight (in grams) of gypsum and Nickel-ammonia coordination compound thus produced is ____.
(Atomic weights in: H=1, N=14, O=16, S=32, Cl=35.5, Ca=40, Ni=59)

Answer 1

The molecular formula of Gypsum can be chemically represented as $CaS{O_4}.2{H_2}O$
Formula Used:
$total{\text{ }}mass{\text{ }}formed{\text{ }} = {\text{ }}\left( {number{\text{ }}of{\text{ }}moles} \right){\text{ }}\left( {molecular{\text{ }}weight} \right)$

Complete step by step answer:
-The preparation of ammonia can be done by using compounds like ammonium sulphate and calcium hydroxide. This reaction between these two compounds not only results in the formation of Ammonia, but also creates a by-product with the chemical formula: $CaS{O_4}.2{H_2}O$. This is also known as gypsum.
-Now, when the ammonia produced as said above, reacts with $NiC{l_2}.6{H_2}O$, it forms a nickel-ammonia metal complex along with water.
Now both these reactions can be represented as follows:
${(N{H_4})_2}S{O_4} + Ca{(OH)_2} \to CaS{O_4}.2{H_2}O + 2N{H_3}$
$NiC{l_2}.6{H_2}O + 6N{H_3} \to [Ni{(N{H_3})_6}]C{l_2} + 6{H_2}O$
The molecular weights of the following compounds are as follows:
${(N{H_4})_2}S{O_4}$= 2(N) + 8(H) + 1(S) + 4(O) = 2(14) +8(1) +1(32) +4(16) = 132g/mol
$CaS{O_4}.2{H_2}O$ = 1(Ca) + 1(S) + 4(H) + 6(O) = 1(40) +1(32) +4(1) +6(16) = 172g/mol
$NiC{l_2}.6{H_2}O$= 1(Ni) + 2(Cl) + 12(H) + 6(O) = 1(59) +2(35.5) +12(1) +6(16) = 238 g/mol
$[Ni{(N{H_3})_6}]C{l_2}$ = 1(Ni) + 6(N) +2(Cl) + 18(H) = 1(59) +6(14) +2(35.5) +18(1) = 232 g/mol
-According to the question, 1584g of ${(N{H_4})_2}S{O_4}$ is used. Since molecular weight of ${(N{H_4})_2}S{O_4}$ is 132g/mol, the number of moles of ${(N{H_4})_2}S{O_4}$ used are $\dfrac{{1584}}{{132}} = 12mol$
-According to the law of proportion of mass, the number of sulphate ions will be constant on both sides (since it is common between both the reactant and product) 12 moles of gypsum are formed as well.
Hence the total weight of gypsum = (number of moles) (molecular weight)
= (12) (172)
= 2064 g
Hence, 2064 g of gypsum is formed.
Similarly, 952g of $NiC{l_2}.6{H_2}O$ are used. Therefore, the number of moles of $NiC{l_2}.6{H_2}O$ used are
$= \dfrac{{952}}{{238}} = 4.0 mol$
Again, by the law of proportion of masses, the number of moles of Ni will remain constant on both sides. Hence, the number of moles of $[Ni{(N{H_3})_6}]$ produced will also be 4.0 moles.
Hence $the{\text{ }}total{\text{ }}mass{\text{ }}of[Ni{(N{H_3})_6}]C{l_2}formed{\text{ }} = {\text{ }}\left( {number{\text{ }}of{\text{ }}moles} \right){\text{ }}\left( {molecular{\text{ }}weight} \right)$
$\; = {\text{ }}\left( 4 \right){\text{ }}\left( {232} \right)$
= 928g

Hence the combined weight (in grams) of gypsum and Nickel-ammonia coordination compound thus produced is = $2064{\text{ }} + {\text{ }}928{\text{ }} = {\text{ }}2992{\text{ }}g$.

Note:
The formation of the products depends on the amount of reaction that has been completed. Suppose in a reaction, only 85% of the reaction has been completed and you are asked to find the amount of product formed, then you must follow all the steps as mentioned in this question and then multiply the final answer by 85%, i.e. 0.85.