Constructing Equations from Reacting Masses and Gas Volumes

Stoichiometry and Molar Quantities

A balanced chemical equation provides the molar quantities of the substances involved in a reaction. For instance, the equation for the reaction of magnesium with oxygen, $Mg(s) + \frac{1}{2}O_2(g) \rightarrow MgO(s)$, indicates that 1 mole of magnesium reacts with 0.5 moles of oxygen gas to produce 1 mole of magnesium oxide. This fundamental relationship allows chemists to work backwards: if we know the experimental masses or volumes of the substances that reacted, we can calculate their molar ratios to determine the balanced equation.

Constructing Equations from Gas Volume Data

According to the properties of an ideal gas, the volume occupied by a gas depends on the temperature and pressure rather than the identity of the gas. Therefore, equal volumes of different gases, measured under the same conditions, contain the same number of moles. This simplifies the construction of equations for gaseous reactions, as the volume ratio of the reacting gases is directly equal to the ratio of their stoichiometric coefficients.

Worked Example: Combustion of a Gaseous Alkane

Suppose 100 $cm^3$ of a gaseous alkane requires 500 $cm^3$ of oxygen for complete combustion, with all volumes measured at the same temperature and pressure. To find the equation:

1. Determine the volume ratio: The ratio of alkane to oxygen is $100 : 500$, which simplifies to $1 : 5$.

2. Apply Avogadro's law: This volume ratio is the mole ratio. Therefore, 1 mole of the alkane reacts with 5 moles of $O_2$.

3. Use the general alkane formula: Let the alkane be $C_n H_{2n+2}$. The unbalanced equation is $C_n H_{2n+2} + 5O_2 \rightarrow nCO_2 + (n+1)H_2O$.

4. Balance the oxygen atoms: The total number of oxygen atoms in the products is $2n$ (from $CO_2$) plus $n + 1$ (from $H_2O$), which equals $3n + 1$. On the reactant side, we have $5 \times 2 = 10$ oxygen atoms.

5. Solve for $n$: $3n + 1 = 10 \Rightarrow 3n = 9 \Rightarrow n = 3$. The alkane is propane, $C_3H_8$. The balanced equation is $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$.

Constructing Equations from Reacting Masses

When dealing with solids or liquids, we use the mass and the relative molar mass ($M_r$) to find the number of moles. The steps are as follows:

1. Calculate the number of moles of each substance using $n = \text{mass} / M_r$.
2. Divide each molar amount by the smallest calculated value to find the simplest ratio.
3. If necessary, multiply the results to obtain the smallest whole number integers.
4. Place these integers as coefficients in the chemical equation.

Worked Example: Neutralisation of a Metal Hydroxide

In an experiment, 1.12 g of a metal hydroxide, $QOH$ (with $M_r = 56$), is required to exactly neutralise a solution containing 0.98 g of sulfuric acid, $H_2SO_4$.

1. Calculate the moles of the hydroxide: $n(QOH) = 1.12 / 56 = 0.020$ mol.

2. Calculate the moles of the acid: First, find the $M_r$ of $H_2SO_4$. $M_r = (2 \times 1.0) + 32.0 + (4 \times 16.0) = 98.0$. Then, $n(H_2SO_4) = 0.98 / 98.0 = 0.010$ mol.

3. Determine the ratio: The molar ratio of $QOH$ to $H_2SO_4$ is $0.020 : 0.010$, which simplifies to $2 : 1$.

4. Construct the equation: The stoichiometric coefficients are 2 for the hydroxide and 1 for the acid. $2QOH(aq) + H_2SO_4(aq) \rightarrow Q_2SO_4(aq) + 2H_2O(l)$.