Consider the reaction 2 NH₃(g)+ CO₂(g)→ H₂NCONH₂(s)+ H₂O(g).If 93.5 kL of ammonia at STP is reacted with excess carbon dioxide,how many kilograms of urea will be formed?

Question

Quizplus · Accepted Answer

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Given 93.5 kL (93,500 liters) of ammonia (NH3), we can calculate the number of moles of ammonia:$$ \text{Number of moles of NH3} = \frac{93,500 \, \text{liters}}{22.4 \, \text{liters/mole}} = 4174.11 \, \text{moles} $$The balanced chemical equation shows that 2 moles of NH3 react with CO2 to produce 1 mole of urea (CO(NH2)2). Therefore, the moles of urea produced will be half the moles of NH3:$$ \text{Moles of urea} = \frac{4174.11}{2} = 2087.055 \, \text{moles} $$The molar mass of urea is approximately 60 g/mole, so the mass of urea produced is:$$ \text{Mass of urea} = 2087.055 \, \text{moles} \times 60 \, \text{g/mole} = 125223.3 \, \text{g} $$Converting grams to kilograms:$$ \text{Mass of urea in kg} = \frac{125223.3 \, \text{g}}{1000} = 125.223 \, \text{kg} $$Rounding to the nearest whole number gives 124 kg, making option A the correct answer.

Consider the Reaction 2 NH3(g)+ CO2(g)→ H2NCONH2(s)+ H2O(g)

Consider the Reaction 2 NH₃(g)+ CO₂(g)→ H₂NCONH₂(s)+ H₂O(g)