## Avogadro’s Law Revisited

Contents

Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.

We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure.

For example, since nitrogen and hydrogen gases react to produce ammonia gas according to \({\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{0.4em}{0ex}}⟶\phantom{\rule{0.4em}{0ex}}2{\text{NH}}_{3}\left(g\right),\) a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.

The explanation for this is illustrated in the figure below. According to Avogadro’s law, equal volumes of gaseous N_{2}, H_{2}, and NH_{3}, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N_{2} reacts with three molecules of H_{2} to produce two molecules of NH_{3}, the volume of H_{2} required is three times the volume of N_{2}, and the volume of NH_{3} produced is two times the volume of N_{2}.

## Example

### Reaction of Gases

Propane, C_{3}H_{8}(*g*), is used in gas grills to provide the heat for cooking. What volume of O_{2}(*g*) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.

### Solution

The ratio of the volumes of C_{3}H_{8} and O_{2} will be equal to the ratio of their coefficients in the balanced equation for the reaction:

\(\begin{array}{l}{\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)\text{ }\text{ }⟶\text{ }\text{ }3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(l\right)\\ \text{1 volume}+\text{5 volumes}\phantom{\rule{3.5em}{0ex}}\text{3 volumes}+\text{4 volumes}\end{array}\)

From the equation, we see that one volume of C_{3}H_{8} will react with five volumes of O_{2}:

\(2.7\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{8}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{5 L}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}}{1\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{8}}}\phantom{\rule{0.2em}{0ex}}=\text{13.5 L}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}\)

A volume of 13.5 L of O_{2} will be required to react with 2.7 L of C_{3}H_{8}.

## Example

### Volumes of Reacting Gases

Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H_{2}(*g*), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N_{2}?

\({\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{NH}}_{3}\left(g\right)\)

### Solution

Because equal volumes of H_{2} and NH_{3} contain equal numbers of molecules and each three molecules of H_{2} that react produce two molecules of NH_{3}, the ratio of the volumes of H_{2} and NH_{3} will be equal to 3:2. Two volumes of NH_{3}, in this case in units of billion ft^{3}, will be formed from three volumes of H_{2}:

\(683\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{billion}\phantom{\rule{0.2em}{0ex}}{\text{ft}}^{3}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{3 billion}\phantom{\rule{0.2em}{0ex}}{\text{ft}}^{3}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}}{2\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{billion}\phantom{\rule{0.2em}{0ex}}{\text{ft}}^{3}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}}}\phantom{\rule{0.2em}{0ex}}=1.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{billion}\phantom{\rule{0.2em}{0ex}}{\text{ft}}^{3}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\)

The manufacture of 683 billion ft^{3} of NH_{3} required 1020 billion ft^{3} of H_{2}. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)

## Example

### Volume of Gaseous Product

What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?

\(2\text{Ga}\left(s\right)+6\text{HCl}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{GaCl}}_{3}\left(aq\right)+3{\text{H}}_{2}\left(g\right)\)

### Solution

To convert from the mass of gallium to the volume of H_{2}(*g*), we need to do something like this:

The first two conversions are:

\(8.88\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{g Ga}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol Ga}}}{69.723\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{g Ga}}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{3 mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}}{2\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol Ga}}}\phantom{\rule{0.2em}{0ex}}=0.191\phantom{\rule{0.2em}{0ex}}{\text{mol H}}_{2}\)

Finally, we can use the ideal gas law:

\({V}_{{\text{H}}_{2}}={\left(\cfrac{nRT}{P}\right)}_{{\text{H}}_{2}}=\phantom{\rule{0.2em}{0ex}}\cfrac{0.191\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{0.08206 L}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{atm}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−1}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{300 K}}{0.951\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{atm}}}\phantom{\rule{0.2em}{0ex}}=\text{4.94 L}\)

### Susan Solomon

Atmospheric and climate scientist Susan **Solomon** (see the picture below) is the author of one of *The New York Times* books of the year (*The Coldest March*, 2001), one of Time magazine’s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.