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Chemical stoichiometry and gases

Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.

We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.

Avogadro’s law revisited

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 N 2 ( g ) + 3 H 2 ( g ) 2 NH 3 ( g ) , 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 [link] . 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 .

This diagram provided models the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3 H subscript 2 followed by an arrow pointing right to N H subscript 3. Just above the formulas, space-filling models are provided. Above N H subscript 2, two blue spheres are bonded. Above 3 H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above N H subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon which is labeled “N subscript 2”. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled “H subscript 2.” Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows which points right to two light green balloons which are each labeled “N H subscript 3.” Each light green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.
One volume of N 2 combines with three volumes of H 2 to form two volumes of NH 3 .

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:

C 3 H 8 ( g ) + 5 O 2 ( g ) 3 CO 2 ( g ) + 4 H 2 O ( l ) 1 volume + 5 volumes 3 volumes + 4 volumes

From the equation, we see that one volume of C 3 H 8 will react with five volumes of O 2 :

2.7 L C 3 H 8 × 5 L O 2 1 L C 3 H 8 = 13.5 L 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 .

Check your learning

An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C 2 H 2 , at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 × 10 3 L of O 2 at 0 °C and 1 atm, will be required to burn the acetylene?

2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O

Answer:

3.34 tanks (2.34 × 10 4 L)

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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