Calculate the moles of urea formed from 340 kg of ammonia and excess carbon dioxide.

$$\text{Molar mass of NH}_3 = \text{Atomic mass of N} + (3 \times \text{Atomic mass of H})$$ $$\text{Molar mass of NH}_3 = 14.0 \text{ g/mol} + (3 \times 1.0 \text{ g/mol})$$ $$\text{Molar mass of NH}_3 = 14.0 \text{ g/mol} + 3.0 \text{ g/mol} = 17.0 \text{ g/mol}$$

$$\text{Moles of NH}_3 = \frac{\text{Mass of NH}_3}{\text{Molar mass of NH}_3}$$ $$\text{Moles of NH}_3 = \frac{340000 \text{ g}}{17.0 \text{ g/mol}}$$ $$\text{Moles of NH}_3 = 20000 \text{ mol}$$

The balanced chemical equation is:

$$2\text{NH}_3\text{(g)} + \text{CO}_2\text{(g)} \rightarrow \text{(NH}_2)_2\text{CO(aq)} + \text{H}_2\text{O(l)}$$

From the equation, 2 moles of $\text{NH}_3$ produce 1 mole of urea ($\text{(NH}_2)_2\text{CO}$).
Since $\text{CO}_2$ is in excess, $\text{NH}_3$ is the limiting reactant.

$$\text{Moles of urea} = \text{Moles of NH}_3 \times \frac{1 \text{ mol urea}}{2 \text{ mol NH}_3}$$ $$\text{Moles of urea} = 20000 \text{ mol} \times \frac{1}{2}$$ $$\text{Moles of urea} = 10000 \text{ mol}$$

The moles of urea that were formed are $\text{10000 mol}$.