Question

missed this? watch kcv: limiting reactant, theoretical yield, and percent yield. we - limiting reactant and theoretical yield: read section 4.4. you can click on the review link to access the section in your e - text. urea (ch₄n₂o) is a common fertilizer that is synthesized by the reaction of ammonia (nh₃) with carbon dioxide. 2nh₃(aq)+co₂(aq)→ch₄n₂o(aq)+h₂o(l) 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. part a determine the limiting reactant. express your answer as a chemical formula. view available hint(s) part b complete previous part(s) part c complete previous part(s) provide feedback

Explanation:

Step1: Calculate moles of reactants

The molar mass of $NH_3$ is $M_{NH_3}=14 + 3\times1=17\ g/mol$. The mass of $NH_3$ is $m_{NH_3}=145.7\times10^{3}\ g$. So the number of moles of $NH_3$, $n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{145.7\times 10^{3}\ g}{17\ g/mol}=8570.59\ mol$.
The molar mass of $CO_2$ is $M_{CO_2}=12 + 2\times16 = 44\ g/mol$. The mass of $CO_2$ is $m_{CO_2}=231.1\times10^{3}\ g$. So the number of moles of $CO_2$, $n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{231.1\times 10^{3}\ g}{44\ g/mol}=5252.27\ mol$.

Step2: Use mole - ratio from balanced equation

The balanced chemical equation is $2NH_3(aq)+CO_2(aq)
ightarrow CH_4N_2O(aq)+H_2O(l)$. The mole - ratio of $NH_3$ to $CO_2$ is $2:1$.
For every 1 mole of $CO_2$, 2 moles of $NH_3$ are required.
If all $n_{CO_2}=5252.27\ mol$ of $CO_2$ reacts, the moles of $NH_3$ required are $n_{NH_3,required}=2\times n_{CO_2}=2\times5252.27\ mol = 10504.54\ mol$. But we have only $n_{NH_3}=8570.59\ mol$.
If all $n_{NH_3}=8570.59\ mol$ of $NH_3$ reacts, the moles of $CO_2$ required are $n_{CO_2,required}=\frac{n_{NH_3}}{2}=\frac{8570.59\ mol}{2}=4285.295\ mol$. Since we have more $CO_2$ than required when all $NH_3$ is consumed, $NH_3$ is the limiting reactant.

Answer:

$NH_3$