Consider the reaction 2 NH₃(g)+ CO₂(g)$\rightarrow$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, 1 mole of gas occupies 22.4 L. Therefore, 93.5 kL of NH3 is equivalent to 93,500 L / 22.4 L/mol = 4174.11 moles of NH3. The reaction shows that 2 moles of NH3 produce 1 mole of urea. Thus, 4174.11 moles of NH3 will produce 2087.055 moles of urea. The molar mass of urea (H2NCONH2) is 60.06 g/mol. Therefore, the mass of urea formed is 2087.055 moles × 60.06 g/mol = 125,364.3 g = 125.364 kg, which is closest to 124 kg.

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

Consider the Reaction 2 NH₃(g)+ CO₂(g) $\rightarrow$ H₂NCONH₂(s)+ H₂O(g)