Answer
Verified
394.2k+ views
Hint: 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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE