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Question: Urea \(\left( {N{H_2}CON{H_2}} \right)\) is formed when \(C{O_2}\) reacts with \(N{H_3}\). \(C{O_2} ...

Urea (NH2CONH2)\left( {N{H_2}CON{H_2}} \right) is formed when CO2C{O_2} reacts with NH3N{H_3}. CO2+2NH3NH2CONH2+H2OC{O_2} + 2N{H_3} \to N{H_2}CON{H_2} + {H_2}O
NH3N{H_3} and CO2C{O_2} are obtained in the following step:
I:N2+3H290%2NH3I:{N_2} + 3{H_2}\xrightarrow{{90\% }}2N{H_3}
II:CaCO350%CaO+CO2II:CaC{O_3}\xrightarrow{{50\% }}CaO + C{O_2}
The Amounts of N2{N_2} and CaCO3CaC{O_3} required to produce 60kgs60kgs of urea are Xkg'X'kg and Ykg'Y'kg respectively. The value of X+YX + Y is:

Explanation

Solution

Urea is known as carbamide. It is the diamide form of carbonic acid. It is the chief product of the metabolic protein breakdown in all the mammals and aquatic animals. It does not only occur in urine but also in blood, bile and sweat of mammals.

Complete step by step answer:
The reaction of formation of urea is given as follows:
CO2+2NH3NH2CONH2+H2OC{O_2} + 2N{H_3} \to N{H_2}CON{H_2} + {H_2}O
The reactant of urea are produced from the following steps:
I:N2+3H290%2NH3I:{N_2} + 3{H_2}\xrightarrow{{90\% }}2N{H_3}
II:CaCO350%CaO+CO2II:CaC{O_3}\xrightarrow{{50\% }}CaO + C{O_2}
Consider the amount of N2{N_2} required to produce 60kgs60kgs of urea to be Xkg'X'kg .
Let the amount of CaCO3CaC{O_3} required to produce 60kgs60kgs of urea be Ykg'Y'kg .
A.First we will calculate the molar masses of N2{N_2},CaCO3CaC{O_3} and NH2CONH2N{H_2}CON{H_2} respectively.
Given data: atomic mass of nitrogen=14 = 14
Atomic mass of oxygen =16 = 16
Atomic mass of hydrogen =1 = 1
Atomic mass of calcium =40 = 40
Atomic mass of carbon =12 = 12
To find : molar mass of N2{N_2} ,CaCO3CaC{O_3} and NH2CONH2N{H_2}CON{H_2}.
Formula to be used: M=m×nM = \sum m \times n
M=M = Molar mass, m=\sum {m = } sum of atomic masses of each element, n=n = number of atoms
Soln:
1.Molar mass of N2={N_2} = atomic mass of nitrogen ×2 \times 2
Substituting the value we get,
Molar mass of N2=14×2{N_2} = 14 \times 2
Molar mass of N2=28g{N_2} = 28g
2.Molar mass of CaCO3=CaC{O_3} = atomic mass of calcium ++ atomic mass of carbon ++ atomic mass of oxygen ×3 \times 3
Substituting the value we get,
Molar mass of CaCO3=40+12+(3×16)CaC{O_3} = 40 + 12 + \left( {3 \times 16} \right)
Molar mass of CaCO3=100gCaC{O_3} = 100g
3.Molar mass of NH2CONH2=N{H_2}CON{H_2} = atomic mass of nitrogen ×2+\times 2 + atomic mass of hydrogen ×4+\times 4 + atomic mass of oxygen ++ atomic mass of carbon
Substituting the value we get,
Molar mass of NH2CONH2=(14×2)+(4×1)+16+12N{H_2}CON{H_2} = \left( {14 \times 2} \right) + \left( {4 \times 1} \right) + 16 + 12
Molar mass of NH2CONH2=28+4+16+12N{H_2}CON{H_2} = 28 + 4 + 16 + 12
Molar mass of NH2CONH2=60gN{H_2}CON{H_2} = 60g
Hence the molar masses of N2{N_2} , CaCO3CaC{O_3} and NH2CONH2N{H_2}CON{H_2}are 28g,100g,60g28g,100g,60g respectively.
B.In the reaction,
One mole of N2{N_2} , CaCO3CaC{O_3} reacts to give one mole of NH2CONH2N{H_2}CON{H_2} .
Which means 28g28g of N2{N_2} and 100g100g of CaCO3CaC{O_3} react to give 60g60g of urea
In terms of kgkg,28kg28kg ofN2{N_2} and 100kg100kg of CaCO3CaC{O_3} react to give 60kg60kg of urea which is comparatively less.
Thus the yield is less than 100%100\% .
We will use the following formula to calculate the mass of nitrogen and calcium carbonate.
%=WMW×100\% = \dfrac{W}{{MW}} \times 100 where, %=\% = percentage, W=W = weight, MW=MW = molecular weight
a.I:N2+3H290%2NH3I:{N_2} + 3{H_2}\xrightarrow{{90\% }}2N{H_3},
In this reaction 90%90\% of nitrogen reacted to give ammonia,
Using the above formula we get,
%\% of nitrogen =WMW×100 = \dfrac{W}{{MW}} \times 100
Rearranging the formula we get,
Weight of nitrogen =  %×MW100 = \dfrac{{\;\% \times MW}}{{100}}
Substituting the value we get,
Weight of nitrogen =90×28100 = \dfrac{{90 \times 28}}{{100}}
Weight of nitrogen =25.2kg = 25.2kg
b.II:CaCO350%CaO+CO2II:CaC{O_3}\xrightarrow{{50\% }}CaO + C{O_2}
In this reaction 50%50\% of nitrogen reacted to give ammonia,
Using the above formula we get,
%\% of calcium carbonate =WMW×100 = \dfrac{W}{{MW}} \times 100
Rearranging the formula we get,
Weight of calcium carbonate =  %×MW100 = \dfrac{{\;\% \times MW}}{{100}}
Substituting the value we get,
Weight of nitrogen =50×100100 = \dfrac{{50 \times 100}}{{100}}
Weight of nitrogen =50kg = 50kg

The Amounts N2{N_2}and CaCO3CaC{O_3} required to produce 60kgs60kgs of urea are 25.2kg25.2kg and 50kg50kg respectively.
Therefore the value of X+Y=75.2kgX + Y = 75.2kg .

Note: Urea is a crystalline compound that consists of 46%46\% nitrogen in the dry state. Urea in the concentrated form forms urine. It is a crystalline compound and is produced in the high pressure and high temperature reactor.