Stoichiometry Figure 2 is the quantitative study of the relative amounts of reactants and products in chemical reactions. Gas stoichiometry is the special case involving 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.
Gas stoichiometry calculations solve for the unknown volume or mass of a gaseous product or reactant. The gases consumed or produced are typically approximated as ideal, meaning point masses that interact solely through perfectly elastic collisions. Perfectly elastic collisions are those in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter Figure 1. Through this approximation, the ideal gas law can be employed to solve for unknown masses or volumes of reactants or products in these reactions. Gas stoichiometry is the quantitative relationship (ratio) between reactants and products in a chemical reaction with reactions that produce gases. Gas stoichiometry applies when the gases produced are assumed to be ideal, and the temperature, pressure, and volume of the gases are all known. Often, but not always, the standard temperature and pressure (STP) are taken as 0 °C and 1 bar and used as the conditions for gas stoichiometric calculations.
A typical problem will give a chemical equation and ask how many grams of one gas will be consumed by reacting with a certain number of grams of another gas. To evaluate the amount of gas consumed or produced in a reaction, it is necessary to start with a balanced chemical reaction. Then, the relationship between different reactants and products can be expressed as a ratio. However, a chemical equation's stoichiometry is given in terms of moles, not grams. Therefore, the ideal gas law must be employed to solve the volume or mass of gas involved in the reaction.
For example, if we wanted to calculate the volume of gaseous NO2 produced from the combustion of 100 g of NH3, by the reaction: 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (l), we would carry out the following calculations:
100g NH3 x 1 mol NH3/17.034g NH3 = 5.871 mol NH3
However, a chemical equation's stoichiometry is given in terms of moles, not grams. Therefore, the ideal gas law must be employed to solve the volume or mass of gas involved in the reaction. There is a 1:1 molar ratio of NH3 to NO2 in the above balanced combustion reaction, so 5.871 mol of NO2 will be formed. We will employ the ideal gas law to solve for the volume at 0 °C (273.15 K) and 1 atmosphere using the gas law constant of R = 0.08206 L · atm · K−1 · mol−1.
One common expression of the ideal gas law is:
PV = nRT
- P = pressure
- V = gas volume
- n = number of moles
- R = universal ideal gas constant
- T = temperature