Chapter 4, Problem 54E

Billions of pounds of urea, $\text{CO}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{,}$ are produced annually for use as a fertilizer. The principal reaction employed is:

${\text{2NH}}_{\text{3}}{\text{+ CO}}_{\text{2}}\to \text{ CO}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}{\text{+ H}}_{\text{2}}\text{O}$

By assuming unlimited amounts of CO₂, how many moles of urea can be produced from each of the following amounts of ${\text{NH}}_{\text{3}}$?

$\begin{array}{l}\text{a}{\text{. 2 mol NH}}_{\text{3}}\hfill \\ \text{b}\text{. 0}{\text{.45 mol NH}}_{\text{3}}\hfill \\ \text{c}{\text{. 10 g NH}}_{\text{3}}\hfill \\ \text{d}\text{. 2}{\text{.0 kg NH}}_{\text{3}}\hfill \end{array}$

Interpretation Introduction

Interpretation:

The number of moles of urea is to be determined in the given amounts of ${\text{NH}}_{\text{3}}$.

Concept introduction:

The characteristic mass of a compound is the formula mass of the compound, which is obtained by adding the atomic masses of all atoms.

A conversion factor is a numerical ratio, which is used to change one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 54E

Solution: The number of moles of urea present in the molecules is given as follows:

$1\text{mol urea}$

$0.23\text{mol urea}$

$0.3\text{mol urea}$

$59\text{mol urea}$

Explanation of Solution

a) $2{\text{mol NH}}_3$

The equivalence is obtained between moles of ${\text{NH}}_{\text{3}}$ and moles of urea with the help of the chemical formula as follows:

$1\text{mol urea}\equiv 2{\text{mol NH}}_3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2\cancel{{\text{mol NH}}_3}\times \frac{1\text{mol urea}}{2\cancel{{\text{mol NH}}_3}}=1\text{mol urea}$

Therefore, the number of moles of urea is $\text{1 mol}$.

b) $0.45{\text{mol NH}}_3$

The equivalence is obtained between moles of ${\text{NH}}_{\text{3}}$ and moles of urea with the help of the chemical formula as follows:

$1\text{mol urea}\equiv 2{\text{mol NH}}_3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$0.45\cancel{{\text{mol NH}}_3}\times \frac{1\text{mol urea}}{2\cancel{{\text{mol NH}}_3}}=0.23\text{mol urea}$

Therefore, the moles of urea are $\text{0}\text{.23 mol}$.

c) $10{\text{g NH}}_3$

The equivalence is obtained between grams of ${\text{NH}}_{\text{3}}$ and moles of ${\text{NH}}_{\text{3}}$ with the help of the chemical formula as follows:

$1{\text{mol NH}}_{\text{3}}\equiv 17.03{\text{g NH}}_{\text{3}}$

The equivalence is obtained between moles of ${\text{NH}}_{\text{3}}$ and moles of urea with the help of the chemical formula as follows:

$1\text{mol urea}\equiv 2{\text{mol NH}}_3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$\text{10}\cancel{{\text{g NH}}_3}\times \frac{1\cancel{{\text{mol NH}}_3}}{17.03\cancel{{\text{g NH}}_3}}\times \frac{1\text{mol urea}}{2\cancel{{\text{mol NH}}_3}}=0.3\text{mol urea}$

Here,

$\text{molar}{\text{mass of NH}}_3=17.03\text{g/mol}$

Therefore, the moles of urea are $\text{0}\text{.3 mol}$.

d) $2.0{\text{kg NH}}_3$

The equivalence is obtained between grams of ${\text{NH}}_{\text{3}}$ and moles of ${\text{NH}}_{\text{3}}$ with the help of the chemical formula as follows:

$1{\text{mol NH}}_{\text{3}}\equiv 17.03{\text{g NH}}_{\text{3}}$

The equivalence is obtained between moles of ${\text{NH}}_{\text{3}}$ and moles of urea with the help of the chemical formula as follows:

$1\text{mol urea}\equiv 2{\text{mol NH}}_3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$\text{2}\cancel{\text{kg}}\cancel{{\text{NH}}_3}\times \left(\frac{1000\cancel{\text{g}}}{1\cancel{\text{kg}}}\right)\times \left(\frac{1\cancel{{\text{mol NH}}_3}}{17.03\cancel{\text{g}}\cancel{{\text{NH}}_3}}\right)\times \left(\frac{1\text{mol urea}}{2\cancel{{\text{mol NH}}_3}}\right)=59\text{mol urea}$

Here,

$\text{molar}{\text{mass of NH}}_3=17.03\text{g/mol}$

Therefore, the moles of urea are $59\text{ mol}$.