Question

Urea $\left( {N{H_2}CON{H_2}} \right)$ is formed when $C{O_2}$ reacts with $N{H_3}$. $C{O_2} + 2N{H_3} \to N{H_2}CON{H_2} + {H_2}O$
$N{H_3}$ and $C{O_2}$ are obtained in the following step:
$I:{N_2} + 3{H_2}\xrightarrow{{90\% }}2N{H_3}$
$II:CaC{O_3}\xrightarrow{{50\% }}CaO + C{O_2}$
The Amounts of ${N_2}$ and $CaC{O_3}$ required to produce $60kgs$ of urea are $'X'kg$ and $'Y'kg$ respectively. The value of $X + Y$ is:

Answer 1

Hint:
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:
$C{O_2} + 2N{H_3} \to N{H_2}CON{H_2} + {H_2}O$
The reactant of urea are produced from the following steps:
$I:{N_2} + 3{H_2}\xrightarrow{{90\% }}2N{H_3}$
$II:CaC{O_3}\xrightarrow{{50\% }}CaO + C{O_2}$
Consider the amount of ${N_2}$ required to produce $60kgs$ of urea to be $'X'kg$ .
Let the amount of $CaC{O_3}$ required to produce $60kgs$ of urea be $'Y'kg$ .
A.First we will calculate the molar masses of ${N_2}$,$CaC{O_3}$ and \[N{H_2}CON{H_2}\] respectively.
Given data: atomic mass of nitrogen$ = 14$
Atomic mass of oxygen $ = 16$
Atomic mass of hydrogen $ = 1$
Atomic mass of calcium $ = 40$
Atomic mass of carbon $ = 12$
To find : molar mass of ${N_2}$ ,$CaC{O_3}$ and \[N{H_2}CON{H_2}\].
Formula to be used: $M = \sum m \times n$
$M = $ Molar mass, $\sum {m = } $ sum of atomic masses of each element, $n = $ number of atoms
Soln:
1.Molar mass of ${N_2} = $ atomic mass of nitrogen $ \times 2$
Substituting the value we get,
Molar mass of ${N_2} = 14 \times 2$
Molar mass of ${N_2} = 28g$
2.Molar mass of $CaC{O_3} = $ atomic mass of calcium $ + $ atomic mass of carbon $ + $ atomic mass of oxygen $ \times 3$
Substituting the value we get,
Molar mass of $CaC{O_3} = 40 + 12 + \left( {3 \times 16} \right)$
Molar mass of $CaC{O_3} = 100g$
3.Molar mass of $N{H_2}CON{H_2} = $ atomic mass of nitrogen $ \times 2 + $ atomic mass of hydrogen $ \times 4 + $ atomic mass of oxygen $ + $ atomic mass of carbon
Substituting the value we get,
Molar mass of $N{H_2}CON{H_2} = \left( {14 \times 2} \right) + \left( {4 \times 1} \right) + 16 + 12$
Molar mass of $N{H_2}CON{H_2} = 28 + 4 + 16 + 12$
Molar mass of $N{H_2}CON{H_2} = 60g$
Hence the molar masses of ${N_2}$ , $CaC{O_3}$ and \[N{H_2}CON{H_2}\]are $28g,100g,60g$ respectively.
B.In the reaction,
One mole of ${N_2}$ , $CaC{O_3}$ reacts to give one mole of $N{H_2}CON{H_2}$ .
Which means $28g$ of ${N_2}$ and $100g$ of $CaC{O_3}$ react to give $60g$ of urea
In terms of $kg$,$28kg$ of${N_2}$ and $100kg$ of $CaC{O_3}$ react to give $60kg$ of urea which is comparatively less.
Thus the yield is less than $100\% $.
We will use the following formula to calculate the mass of nitrogen and calcium carbonate.
$\% = \dfrac{W}{{MW}} \times 100$ where, $\% = $ percentage, $W = $ weight, $MW = $ molecular weight
a.$I:{N_2} + 3{H_2}\xrightarrow{{90\% }}2N{H_3}$,
In this reaction $90\% $ of nitrogen reacted to give ammonia,
Using the above formula we get,
$\% $ of nitrogen $ = \dfrac{W}{{MW}} \times 100$
Rearranging the formula we get,
Weight of nitrogen $ = \dfrac{{\;\% \times MW}}{{100}}$
Substituting the value we get,
Weight of nitrogen $ = \dfrac{{90 \times 28}}{{100}}$
Weight of nitrogen $ = 25.2kg$
b.$II:CaC{O_3}\xrightarrow{{50\% }}CaO + C{O_2}$
In this reaction $50\% $ of nitrogen reacted to give ammonia,
Using the above formula we get,
$\% $ of calcium carbonate $ = \dfrac{W}{{MW}} \times 100$
Rearranging the formula we get,
Weight of calcium carbonate $ = \dfrac{{\;\% \times MW}}{{100}}$
Substituting the value we get,
Weight of nitrogen $ = \dfrac{{50 \times 100}}{{100}}$
Weight of nitrogen $ = 50kg$

The Amounts ${N_2}$and $CaC{O_3}$ required to produce $60kgs$ of urea are $25.2kg$ and $50kg$ respectively.
Therefore the value of $X + Y = 75.2kg$ .

Note:Urea is a crystalline compound that consists of $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.