Question

1. urea (ch₄n₂o) is a common fertilizer that can be synthesized by the reaction of ammonia (nh₃) with carbon dioxide: 2 nh₃(aq) + co₂(aq) → ch₄n₂o(aq) + h₂o(l) in an industrial synthesis of urea, a chemist combines 136.4 kg of ammonia with 211.4 kg of carbon dioxide and obtains 168.4 kg of urea. determine the limiting reactant, theoretical yield of urea, and percent yield for the reaction.

Explanation:

Step1: Calculate moles of reactants

First, find molar masses: $M_{NH_3}=14 + 3\times1=17\ g/mol$, $M_{CO_2}=12+2\times16 = 44\ g/mol$, $M_{CH_4N_2O}=12 + 4\times1+2\times14+16=60\ g/mol$.
Moles of $NH_3$, $n_{NH_3}=\frac{136.4\times1000\ g}{17\ g/mol}=8023.53\ mol$.
Moles of $CO_2$, $n_{CO_2}=\frac{211.4\times1000\ g}{44\ g/mol}=4804.55\ mol$.

Step2: Determine limiting reactant

From the balanced equation $2NH_3(aq)+CO_2(aq)\to CH_4N_2O(aq)+H_2O(l)$, the mole - ratio of $NH_3$ to $CO_2$ is 2:1.
For $n_{CO_2} = 4804.55\ mol$, the moles of $NH_3$ required is $2\times4804.55\ mol = 9609.1\ mol$. But we have only $8023.53\ mol$ of $NH_3$. So, $NH_3$ is the limiting reactant.

Step3: Calculate theoretical yield of urea

Since $NH_3$ is the limiting reactant, from the mole - ratio of $NH_3$ to $CH_4N_2O$ (2:1), moles of $CH_4N_2O$ formed is $\frac{n_{NH_3}}{2}=\frac{8023.53\ mol}{2}=4011.77\ mol$.
Theoretical mass of $CH_4N_2O$, $m_{theo}=4011.77\ mol\times60\ g/mol = 240706.2\ g=240.71\ kg$.

Step4: Calculate percent yield

Percent yield, $\%yield=\frac{actual\ yield}{theoretical\ yield}\times100=\frac{168.4\ kg}{240.71\ kg}\times100 = 70.0\ \%$.

Answer:

Limiting reactant: $NH_3$
Theoretical yield of urea: $240.71\ kg$
Percent yield: $70.0\ \%$