Gases

Stoichiometry of Gases

Stoichiometry of gases relies on the partial pressure of gases—the pressure of one gas in a mixture of gases.

Stoichiometry is the relationship between the amounts of products and reactants in a reaction. The law of conservation of mass states that the mass of the reactants must be equal to the mass of the products. The ideal gas law leads to the understanding that one mole of any gas occupies 22.4 liters of volume at standard temperature and pressure (STP). Together, these laws allow for the calculation of amounts of products and reactants in a chemical reaction involving gases.

For example, consider the combustion of ammonia: 4NH3(g)+7O2(g)4NO2(g)+6H2O(l)4{\rm{NH}}_{3}(g)+7{\rm O}_{2}( g)\rightarrow4{\rm{NO}}_{2}{(g)}+6{\rm H}_{2}{\rm O}(l). The volume of NO2 gas that is produced from the combustion of 12.0 grams of NH3 can be calculated using the volume of gas at STP.
(12.0gNH31)(1molNH317.04g)(4molNO24molNH3)(22.4L1molNO2)=15.8LNO2\left(\frac{{12.0\;{\rm g}\;{\rm{NH}}_3}}{1}\right)\!\left(\frac{1\;{\rm{mol}}\;{{\rm{NH}}}_3}{17.04\;\rm g}\right)\!\left(\frac{4\;{\rm{mol}}\;{\rm{NO}}_2}{4\;{\rm{mol}}\;{\rm{NH}}_3}\right)\!\left(\frac{22.4\;{\rm L}}{1\;{\rm{mol}}\;{\rm{NO}}_{2}}\right)=15.8\;{\rm L}\;{\rm{NO}}_{2}
In chemical reactions, it is impossible to keep the gases separate from one another. When more than one gas is present in a container, each gas exerts pressure on the container individually. The partial pressure is the pressure of an ideal gas that contributes to the total pressure of a mixture of gases at constant temperature. Dalton's law of partial pressures states that the total pressure (Ptotal) of a mixture of ideal and nonreacting gases is the sum of the partial pressures of the individual gases.
Ptotal=P1+P2+P3+PnP_{\rm{total}}=P_1+P_2+P_3\mathellipsis+P_n
For example, consider a gas cylinder of a mixture of oxygen gas (O2) and nitrogen gas (N2). The partial pressure of O2 is 0.65 atm. The partial pressure of N2 is 0.15 atm. Assuming the mixture behaves as an ideal gas, the total pressure of the gas in the cylinder is the sum of the partial pressures.
Ptotal=PO2+PN2=0.65atm+0.15atm=0.80atm\begin{aligned}P_{\rm{total}}&=P_{\rm{O_2}}+P_{\rm{N_2}}\\ &=0.65\;\rm{atm}+0.15\;\rm{atm}\\ &=0.80\;\rm{atm}\end{aligned}
The amount of a gas present in a mixture of gases can also be described by its mole fraction (χ\chi), the concentration expressed as the moles of solvent divided by the total number of all moles in a solution. For a gas it is the number of moles of gas (i) divided by the total number of moles in the gas mixture.
χ=moles of gasitotal moles of gas in mixture\chi=\frac{\text{moles of gas}\;i}{\text{total moles of gas in mixture}}
The partial pressure of a gas can also be stated in terms of its mole fraction.
Pi=(Ptotal)(χ)P_i=\left(P_{\rm{total}}\right)\!\left(\chi\right)
Consider a mixture of 5.0 moles of hydrogen gas (H2) and 4.0 moles of argon (Ar). The total pressure of the mixture is 4.10 atm. The partial pressure of Ar is the product of the total pressure and the mole fraction of Ar.
PAr=(Ptotal)(χAr)=(Ptotal)(moles Artotal moles)=(4.10atm)(4.0mol9.0mol)=1.8atm\begin{aligned}P_{\rm{Ar}}&=\left(P_{\rm{total}}\right)\!\left(\chi_{\rm{Ar}}\right)\\&= \left(P_{\rm{total}}\right)\!\left(\frac {\text{moles }\rm{Ar}} {\text {total moles}}\right)\\&= \left(4.10\;{\rm{atm}}\right)\!\left(\frac {4.0\;{\rm{mol}}} {9.0\;{\rm {mol}} } \right)\\&= 1.8\;{\rm{atm}}\end{aligned}
Partial pressure also contributes to the way a gas dissolves in a liquid. According to Henry's law, the amount of a gas that dissolves in a certain type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid at a specific temperature. This is represented by the equation C=kPgasC=kP_{\rm{gas}} , where C is the solubility of the given gas in the given solvent (given in M gas/L or mL gas/L), k is Henry's law constant (usually given in M/atm), and Pgas is the partial pressure of the gas (usually given in atm). This equation can be used to find the concentration of a gas dissolved in a liquid. For example, calculate the concentration of carbon dioxide (CO2) in 1.00 L of water (H2O) at a pressure of 3.10 atm and a temperature of 25.0°C. For CO2 in water, k is 3.36×102M/atm3.36\times10^{-2}\;\rm M/\rm{atm} .
C=kPgas=(3.36×102M/atm)(3.10atm)=0.104M\begin{aligned} C&=kP_{\rm{gas}}\\&=\left(3.36\times10^{-2}\,\rm{M/atm}\right)\!(3.10\;\rm{atm})\\&=0.104\;\rm M\end{aligned}