## Gas Stoichiometry

At standard temperature and pressure, one mole of any gas will occupy a volume of 22.4 L.

### Learning Objectives

Calculate volumes of gases consumed/produced in a reaction using gas stoichiometry.

### Key Takeaways

#### Key Points

- At Standard Temperature and Pressure (STP), 1 mole of any gas will occupy a volume of 22.4 L.
- The Ideal Gas Law, along with a balanced chemical equation, can be used to solve for the amount, either in volume or mass, of gas consumed or produced in a chemical reaction.

#### Key Terms

**stoichiometry**: the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions (chemical equations)

Stoichiometry is the quantitative study of the relative amounts of reactants and products in chemical reactions; gas stoichiometry involves chemical reactions that produce gases. Stoichiometry is based on the law of conservation of mass, meaning that the mass of the reactants must be equal to the mass of the products. This assumption can be used to solve for unknown quantities of reactants or products.

### Stoichiometric Calculations Involving Ideal Gases at STP

Stoichiometric calculations involving gases allow us to convert between mass, number of moles, and most importantly, *volume* of gases. The following relationship makes this possible:

- 1 mole of
*any*gas at standard temperature and pressure (273 K and 1 atm) occupies a volume of 22.4 L.

While the above relationship is an estimation, it is a relatively good approximation at STP, and can be used reliably in calculations.

### Example

[latex]4\;NH_3(g)+7\;O_2(g)\rightarrow4\;NO_2(g)+6\;H_2O(l)[/latex]

- According to the above reaction, what volume of NO
_{2}(*g*) is produced from the combustion of 100 g of NH_{3}(*g*), assuming the reaction takes place at standard temperature and pressure? - From the periodic table, we can determine that the molar mass of ammonia, NH
_{3}(*g*), is 17 g/mol, and perform the following stoichiometric calculation: - [latex]\left(\frac{\text{100 g }NH_3}{ }\right)\times\left(\frac{\text{1 mol }NH_3}{\text{17 }g}\right)\times \left(\frac{\text{4 mol }NO_2}{\text{4 mol }NH_3}\right)\times \left(\frac{\text{22.4 }L}{\text{1 mol }NO_2}\right)=\text{132 L }NO_2(g)[/latex]

Note the final conversion factor. Because we are told that the reaction takes place at STP, we can relate volume, 22.4 L, to 1 mol NO_{2}.