Molar Gas Volumes and Stoichiometry for the ESAT

A balanced chemical equation provides the molar quantities of the substances reacting and the products formed. When dealing with gases, we can relate these molar quantities directly to volumes because of the unique physical properties of gases.

The Nature of Ideal Gases

In the study of chemistry, we often assume gases behave 'ideally'. A primary assumption for an ideal gas is that the volume of the actual gas molecules is so small compared to the total volume the gas occupies that it is considered negligible.

Because the volume of the particles themselves is negligible, the total volume of a gas is determined by the conditions of temperature and pressure rather than the chemical identity of the gas. This leads to a fundamental principle: equal volumes of different gases, when measured at the same temperature and pressure, contain the exact same number of moles and molecules.

Molar Volume at Room Temperature and Pressure

When the specific conditions of a gas are known, we can define its molar volume. For the ESAT, you must know the value for room temperature and pressure (rtp). At rtp, one mole of any gas occupies a volume of 24 $dm^3$.

This constant allows us to calculate the amount of gas in a given volume, or calculate the volume a known amount of gas will occupy. The relationship is expressed by the following formula:

$\text{Number of moles of gas} = \frac{\text{volume}}{\text{molar volume}}$

When using this equation, ensure that both the volume and the molar volume are expressed in the same units (either $dm^3$ or $cm^3$).

Calculating Reacting Gas Volumes

Because the molar ratio in an equation corresponds to the volume ratio for gases, we can solve stoichiometry problems involving gases quite simply if the conditions remain constant.

Worked Example: Octane Combustion

A car engine burns octane vapour, $C_8H_{18}$, according to the following equation:

$2C_8H_{18}(g) + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g)$

What is the minimum volume of oxygen required to ensure the complete combustion of 1 $dm^3$ of octane vapour, assuming all volumes are measured at the same temperature and pressure?

Solution: The balanced equation shows that 2 volumes of octane require 25 volumes of oxygen. If we have 1 $dm^3$ of octane (which is half of 2 volumes), we need half of 25 volumes of oxygen:

$\text{Volume of } O_2 = \frac{1}{2} \times 25 = 12.5\ dm^3$

Calculations involving Mass and Gas Volume

Often, you will need to link the mass of a solid or liquid reactant to the volume of a gaseous product. This requires a two step process: converting to moles first, then using the molar volume.

Worked Example: Producing Ammonia

Calculate the mass of ammonium chloride needed to produce 1.20 $dm^3$ of ammonia at rtp in the following reaction:

$2NH_4Cl(s) + Ca(OH)_2(s) \rightarrow 2NH_3(g) + CaCl_2(s) + 2H_2O(g)$

Assume the molar volume at rtp is 24.0 $dm^3$ and $A_r$ values: $H = 1.0, N = 14.0, Cl = 35.5$.

Solution:

1. Find the moles of $NH_3$ produced: $\text{moles} = \frac{1.20}{24.0} = 0.050\ mol$.
2. Use the molar ratio: The equation shows 2 moles of $NH_4Cl$ produce 2 moles of $NH_3$ (a 1:1 ratio). Therefore, 0.050 mol of $NH_4Cl$ is required.
3. Calculate the mass of $NH_4Cl$: $M_r = 14.0 + (1.0 \times 4) + 35.5 = 53.5$. $\text{Mass} = 0.050 \times 53.5 = 2.68\ g$.

Worked Example: Zinc and Sulfuric Acid

What volume of hydrogen is produced when 3.25 g of zinc reacts with excess dilute sulfuric acid at rtp? $A_r$ value: $Zn = 65$.

Solution:

1. Write the equation: $Zn(s) + H_2SO_4(aq) \rightarrow ZnSO_4(aq) + H_2(g)$.
2. Calculate moles of Zn: $\text{moles} = \frac{3.25}{65} = 0.050\ mol$.
3. Use the molar ratio: 1 mole of $Zn$ produces 1 mole of $H_2$. So, 0.050 mol of $H_2$ is formed.
4. Calculate the volume: $\text{Volume} = 0.050 \times 24.0 = 1.20\ dm^3$.

Constructing Equations from Volume Data

You can determine the identity of an unknown reactant by looking at the ratios of the gas volumes involved in its reaction.

Worked Example: Alkane Combustion

100 $cm^3$ of a gaseous alkane required 500 $cm^3$ of oxygen for complete combustion at constant conditions. Determine the equation.

Solution: The ratio of alkane to oxygen is 100:500, which simplifies to 1:5. Thus, 1 mole of alkane reacts with 5 moles of $O_2$. Using the general formula for an alkane, $C_nH_{2n+2}$, we can balance the oxygen atoms. On the left, we have 10 oxygen atoms ($5 \times O_2$). On the right, we form $n$ molecules of $CO_2$ and $(n+1)$ molecules of $H_2O$.

$\text{Total O atoms on right} = 2n + (n + 1) = 3n + 1$ Setting this equal to the 10 atoms on the left: $10 = 3n + 1$. $9 = 3n$, so $n = 3$. The alkane is propane, $C_3H_8$, and the equation is: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$.