Question

in an industrial synthesis of urea, a chemist combines 145.7 kg of ammonia with 231.1 kg of carbon dioxide and obtains 172.1 kg of urea. 2nh₃(aq) + co₂(aq) → ch₄n₂o(aq) + h₂o(l). part b determine the theoretical yield of urea. express your answer in kilograms to four significant figures. part c complete previous part(s)

Explanation:

Step1: Write the balanced chemical equation

$2NH_3(aq)+CO_2(aq)
ightarrow CH_4N_2O(aq) + H_2O(l)$

Step2: Calculate the molar masses

The molar mass of $NH_3$ is $M_{NH_3}=14.01 + 3\times1.01=17.04\ g/mol$. The molar mass of $CO_2$ is $M_{CO_2}=12.01+2\times16.00 = 44.01\ g/mol$, and the molar mass of $CH_4N_2O$ (urea) is $M_{CH_4N_2O}=12.01 + 4\times1.01+2\times14.01+16.00 = 60.06\ g/mol$.

Step3: Determine the limiting - reactant

The number of moles of $NH_3$, $n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{145.7\times10^{3}\ g}{17.04\ g/mol}=8.551\times10^{3}\ mol$. The number of moles of $CO_2$, $n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{231.1\times10^{3}\ g}{44.01\ g/mol}=5.251\times10^{3}\ mol$. From the balanced equation, the mole - ratio of $NH_3$ to $CO_2$ is $2:1$. For $n_{CO_2} = 5.251\times10^{3}\ mol$ of $CO_2$, the required amount of $NH_3$ is $n_{NH_3}^{required}=2\times n_{CO_2}=2\times5.251\times10^{3}\ mol = 10.502\times10^{3}\ mol$. Since $n_{NH_3}=8.551\times10^{3}\ mol<10.502\times10^{3}\ mol$, $NH_3$ is the limiting reactant.

Step4: Calculate the theoretical yield of urea

From the balanced equation, the mole - ratio of $NH_3$ to $CH_4N_2O$ is $2:1$. So the number of moles of urea produced, $n_{CH_4N_2O}=\frac{1}{2}n_{NH_3}=\frac{1}{2}\times8.551\times10^{3}\ mol = 4.2755\times10^{3}\ mol$. The mass of urea (theoretical yield), $m_{CH_4N_2O}=n_{CH_4N_2O}\times M_{CH_4N_2O}=4.2755\times10^{3}\ mol\times60.06\ g/mol=256.8\ kg$.

Answer:

$256.8\ kg$