Question

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 c determine the percent yield for the reaction. express the yield as a percentage to three significant figures. view available hint(s) urea produced by the limiting reactant (nh3) as determined in part a to grams (g). then, convert grams to kilograms (kg) using the conversion factor 1000 g = 1 kg. in summary: molar mass of urea (60.06 g/mol) to convert moles (mol) to grams (g). then, convert grams to kilograms using the conversion factor 1000 g = 1 kg. nch4n2o = 4280.3 mol ch4n2o × 1 mol ch4n2o / 1 mol ch4n2o × 60.06 g ch4n2o / 1 mol ch4n2o × 1 kg / 1000 g = 257.1 kg ch4n

Explanation:

Step1: Write the balanced chemical equation

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

Step2: Calculate the theoretical yield of urea

First, find the moles of $NH_3$ and $CO_2$ from given masses. But we are not given the masses to calculate theoretical yield this way. However, from the given information in part - related text, we assume we have already found the theoretical yield of urea $m_{theoretical}=257.1$ kg (from the shown calculation in the image). The actual yield of urea $m_{actual}=172.1$ kg.

Step3: Calculate the percent - yield

The formula for percent - yield is $\text{Percent Yield}=\frac{m_{actual}}{m_{theoretical}}\times100\%$.
Substitute $m_{actual} = 172.1$ kg and $m_{theoretical}=257.1$ kg into the formula:
$\text{Percent Yield}=\frac{172.1}{257.1}\times100\%$
$=\frac{17210}{257.1}\%$
$\approx 67.0\%$

Answer:

$67.0\%$