Question

4. What mass of urea is produced from 9.0 liters of ammonia?

The reaction is 2NH3(g) + CO2(g) → H2NCONH2(g) + H2Og)

Answer 1

12.07 g.

Step-by-step explanation:

• The balanced equation for the mentioned reaction is:

2NH₃(g) + CO₂(g) → H₂NCONH₂(g) + H₂O(g),

It is clear that 2.0 moles of NH₃ react with 1.0 mole of CO₂ to produce 1.0 mole of H₂NCONH₂ and 1.0 moles of H₂O.

• Consider the reaction proceeds at STP conditions:

At STP, 9.0 L of NH₃ react with an excess of CO₂ gas:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

using cross multiplication:

1.0 mol of NH₃ represents → 22.4 L.

??? mol of NH₃ represents → 9.0 L.

∴ 9.0 L of NH₃ represents = (1.0 mol)(9.0 L)/(22.4 L) = 0.4018 mol.

• To find the no. of moles of urea (H₂NCONH₂) produced:

Using cross multiplication:

2.0 mol of NH₃ produce → 1.0 mol of H₂NCONH₂, from stichiometry.

0.4018 mol of NH₃ produce → ??? mol of H₂NCONH₂.

∴ The no. of moles of H₂NCONH₂ = (1.0 mol)(0.4018 mol)/(2.0 mol) = 0.201 mol.

• Now, we can find the mass of H₂NCONH₂ produced:

mass = n * molar mass = (0.201 mol) * (60.06 g/mol) = 12.07 g.