Gay-Lussac’s Law of Combining Volumes states that when gases react together to form other gases, the volumes of the reactant and product gases (measured at the same temperature and pressure) bear a simple whole-number ratio to one another and to the volumes of the gaseous products .

This means that gaseous volumes combine and produce gases in simple ratios such as 1:1, 1:2, 2:3, etc. These relationships depend on the balanced chemical equation and remain valid at constant temperature and pressure.

Mathematically, if gases A and B react to form gas C: $$aV_A + bV_B \rightarrow cV_C$$ then their volumes are in the ratio: $$V_A : V_B : V_C = a : b : c$$

This law helps in predicting the volume of reactants required or products formed in gaseous reactions under the same physical conditions.

A limiting factor in a chemical reaction is the reactant that is completely used up first during the reaction, thereby determining the amount of product formed. Once the limiting reactant is exhausted, the reaction stops, even if other reactants are still present in excess. The limiting factor thus controls the extent of the reaction.

For example, in a reaction between hydrogen and oxygen to form water, if hydrogen is present in a smaller proportion than required by the balanced equation, it will be the limiting reactant. The other reactant, oxygen, will then be in excess. Identifying the limiting factor helps in calculating the exact yield of products and determining how much of each reactant remains unreacted.

Example 1: Hydrogen reacts with oxygen according to the equation:
$$2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$$ If 100 cm³ of oxygen gas reacts completely, what volume of hydrogen is required and what volume of steam is produced?

Volume ratio: $$H_2 : O_2 : H_2O = 2 : 1 : 2$$

For 1 volume of O₂, 2 volumes of H₂ are needed and 2 volumes of H₂O are produced.

For 100 cm³ of O₂:
$$\text{H}_2 = 2 \times 100 = 200 \, \text{cm}^3$$ $$\text{H}_2O = 2 \times 100 = 200 \, \text{cm}^3$$ Final Answer: Hydrogen used = 200 cm³, Water vapor formed = 200 cm³.

Example 2: Ammonia is formed according to the equation:
$$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$$ If 60 cm³ of nitrogen reacts with 200 cm³ of hydrogen, calculate the volume of ammonia formed and the volume of unused gas.

Volume ratio: $$N_2 : H_2 : NH_3 = 1 : 3 : 2$$ For 1 volume of N₂, 3 volumes of H₂ are required.

For 60 cm³ of N₂:
$$\text{Required } H_2 = 3 \times 60 = 180 \, \text{cm}^3$$ Available H₂ = 200 cm³ → 20 cm³ of H₂ remains unused.

From the ratio, 1 volume of N₂ produces 2 volumes of NH₃:
$$\text{NH}_3 = 2 \times 60 = 120 \, \text{cm}^3$$ Final Answer: Ammonia formed = 120 cm³, Unused hydrogen = 20 cm³.

Example 3: Carbon monoxide reacts with oxygen according to the equation:
$$2CO(g) + O_2(g) \rightarrow 2CO_2(g)$$ If 150 cm³ of CO reacts with 100 cm³ of O₂, determine the volume of CO₂ formed and the gas remaining.

Volume ratio: $$CO : O_2 : CO_2 = 2 : 1 : 2$$ For 2 volumes of CO, 1 volume of O₂ is required.

For 100 cm³ of O₂, required CO = 2 × 100 = 200 cm³ (but only 150 cm³ is available).
Hence, CO is the limiting reagent.

From the ratio, 2 volumes of CO produce 2 volumes of CO₂ → equal ratio.
$$\text{CO}_2 \text{ formed} = 150 \, \text{cm}^3$$ Oxygen used = 75 cm³ → Remaining O₂ = 100 − 75 = 25 cm³.

Final Answer: CO₂ formed = 150 cm³, Unused O₂ = 25 cm³.

Example 4: Hydrogen reacts with oxygen according to the equation:
$$2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$$ If 50 cm³ of hydrogen were sparked with 30 cm³ of oxygen, calculate the volume of unused oxygen after cooling to the initial temperature and pressure.(JAMB)

From the balanced equation the volume ratio is: $$H_2 : O_2 = 2 : 1$$ This means 2 volumes of H₂ react with 1 volume of O₂.

For 50 cm³ of H₂, the required volume of O₂ is: $$\text{Required } O_2 = \frac{1}{2} \times 50 = 25 \,\text{cm}^3$$ Available O₂ = 30 cm³.

Therefore the unused oxygen is: $$\text{Unused } O_2 = 30 - 25 = 5 \,\text{cm}^3$$

Final Answer: Unused O₂ = 5 cm³ (at the initial temperature and pressure).

Note: Cooling to the initial temperature and pressure does not change the amount (moles) of leftover oxygen, so its volume at those conditions remains 5 cm³. If the water formed condenses to liquid on cooling, the only gas remaining is this unused oxygen.

Example 5: When 80 cm³ of carbon monoxide gas reacts with 60 cm³ of oxygen gas according to the equation:
$$2CO(g) + O_2(g) \rightarrow 2CO_2(g)$$ Calculate the volume of excess reactant left after the reaction, assuming all volumes are measured at the same temperature and pressure.

From the balanced equation, the volume ratio is: $$CO : O_2 = 2 : 1$$ That means 2 volumes of CO react with 1 volume of O₂.

For 80 cm³ of CO, the required volume of O₂ is: $$\text{Required } O_2 = \frac{1}{2} \times 80 = 40 \,\text{cm}^3$$ Available O₂ = 60 cm³

Therefore, oxygen is in excess, and carbon monoxide is the limiting reactant.

Volume of excess O₂ left after reaction: $$\text{Excess } O_2 = 60 - 40 = 20 \,\text{cm}^3$$

Final Answer: Excess oxygen = 20 cm³ (at the same temperature and pressure).

• Always write and balance the chemical equation first.
• The ratio of gas volumes directly corresponds to the ratio of coefficients in the balanced equation.
• The limiting gas is the one that is completely used up first during the reaction.
• To find the limiting gas, compare the available volume ratio to the required ratio from the equation.
• Once the limiting gas is identified, use it to calculate the volume of the product formed.
• Subtract the volume of gas used from the initial volume to find any unused gas.
• All gas volumes must be measured under the same temperature and pressure conditions.