Urea is made by heating ammonia and carbon dioxide in a sealed reactor: 2 NH₃ + CO₂ → CO(NH₂)₂ +…

Moles. $n(\text{NH}_3) = 68.0 / 17.0 = 4.00\ \text{mol}$ and $n(\text{CO}_2) = 132.0 / 44.0 = 3.00\ \text{mol}$. Divide by the coefficients: $4.00 / 2 = 2.00$ for ammonia and $3.00 / 1 = 3.00$ for carbon dioxide. Ammonia gives the smaller number, so it is the limiting reagent.

(a) $2\,\text{NH}_3$ gives $1$ urea, so $n(\text{urea}) = 2.00\ \text{mol}$.

(b) $2.00 \times 60.0 = 120.0\ \text{g}$.

(c) $\text{CO}_2$ used is $2.00 \times 44.0 = 88.0\ \text{g}$, so $132.0 - 88.0 = 44.0\ \text{g}$ is left.

(d) Percentage yield $= \dfrac{93.6}{120.0} \times 100 = 78.0\%$.

(e) The urea formed is $93.6 / 60.0 = 1.56\ \text{mol}$. That used $2 \times 1.56 \times 17.0 = 53.04\ \text{g}$ of $\text{NH}_3$ and $1.56 \times 44.0 = 68.64\ \text{g}$ of $\text{CO}_2$. What is left is $(68.0 - 53.04) + (132.0 - 68.64) = 14.96 + 63.36 = 78.3\ \text{g}$.