Q.If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

The key idea is Gay-Lussac's Law of Gaseous Volumes, which relates the volumes of gaseous reactants and products in a simple whole-number ratio under constant temperature and pressure.

1. The balanced chemical equation for the reaction of dihydrogen with dioxygen to form water vapour is:

2. According to Gay-Lussac's Law, the stoichiometric coefficients represent the relative volumes of the reacting gases and gaseous products. Thus, 2 volumes of $\text{H}_2$ react with 1 volume of $\text{O}_2$ to produce 2 volumes of $\text{H}_2\text{O}$. …

When gases react, their volumes behave in a very predictable way, provided the temperature and pressure remain constant. This predictability is captured by Gay-Lussac's Law of Gaseous Volumes. This law states that when gases combine or are produced in a chemical reaction, they do so in simple whole-number ratios by volume.

The fundamental reason this works is Avogadro's Law, which tells us that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. This means that the volume ratio of reacting gases is directly equivalent to their mole ratio, which is given by the coefficients in a balanced chemical equation. Therefore, to solve this problem, we first need the balanced chemical equation for the reaction between dihydrogen and dioxygen to form water vapour.

1. Write the balanced chemical equation for the reaction. Dihydrogen ($H_2$) reacts with dioxygen ($O_2$) to form water ($H_2O$). Since we are dealing with volumes of gases, we consider water in its gaseous state (water vapour). The unbalanced equation is: $H_2(g) + O_2(g) \rightarrow H_2O(g)$ To balance it, we need two hydrogen atoms on both sides and two oxygen atoms on both sides.

This equation tells us that 2 molecules of dihydrogen react with 1 molecule of dioxygen to produce 2 molecules of water vapour.

2. Interpret the balanced equation in terms of volumes.

According to Gay-Lussac's Law and Avogadro's Law, the stoichiometric coefficients in the balanced equation directly represent the volume ratios of the reacting gases and gaseous products.

So, from the balanced equation:

2 volumes of $H_2$ react with 1 volume of $O_2$ to produce 2 volumes of $H_2O$.

3. Determine the limiting reactant (if any) and calculate the product volume.

   We are given:

   • Volume of dihydrogen ($H_2$) = 10 volumes
   • Volume of dioxygen ($O_2$) = 5 volumes

   Let's see how much dioxygen is needed to react completely with 10 volumes of dihydrogen:

   From the ratio $H_2 : O_2 = 2 : 1$, for 10 volumes of $H_2$, we would need: …

Method: Gay-Lussac’s Law of Gaseous Volumes (Volume–Volume Analysis)

This law states that when gases react, they do so in volumes that bear a simple whole-number ratio to each other and to the volumes of products (if gaseous), provided all volumes are measured at the same temperature and pressure.

Steps

1. Write the balanced chemical equation

2. Interpret the volume ratio from the equation

   From the coefficients:

   • 2 volumes of $H_2$ react with 1 volume of $O_2$ to give 2 volumes of $H_2O$ vapour.

3. Apply the given volumes

   Given: 10 volumes of $H_2$ and 5 volumes of $O_2$.

   Check the ratio:

   $\frac{\text{Volume of } H_2}{\text{Volume of } O_2} = \frac{10}{5} = 2$ …

Here are the common mistakes students make on this question, along with how to avoid each — explained concept-first for clarity.

Mistake 1: Forgetting Gay-Lussac’s Law of Gaseous Volumes

The mistake:

Students treat the reaction like a mass-based calculation (using moles and grams) instead of using the volume ratio directly.

Why it’s wrong:

Gay-Lussac’s law states that when gases react at the same temperature and pressure, their volumes are in simple whole-number ratios — same as the coefficients in the balanced equation.

How to avoid:

Always write the balanced chemical equation first:

From the equation:

• 2 volumes of $H_2$ react with 1 volume of $O_2$ to give 2 volumes of $H_2O$ vapour.

So the volume ratio is 2 : 1 : 2.

Mistake 2: Using the Given Volumes Incorrectly

The mistake:

Students directly multiply or divide the given volumes (10 and 5) without checking which reactant is limiting.

Why it’s wrong:

The ratio from the equation is 2:1. Here, 10 volumes of $H_2$ and 5 volumes of $O_2$ are exactly in the 2:1 ratio — so both are fully consumed. But if the ratio were different, one reactant would be in excess.

How to avoid:

• Compare the given volumes to the required ratio.
• Here: $10 : 5 = 2 : 1$ → perfect match.
• So no limiting reagent issue — both react completely.

Mistake 3: Forgetting That Water Vapour Is Also a Gas

The mistake:

Students calculate the volume of liquid water instead of water vapour.

Why it’s wrong:

The question asks for water vapour — a gas at the reaction temperature. Gay-Lussac’s law applies only to gases.

How to avoid:

• Check the physical state in the equation: $(g)$ means gas.
• Volume of water vapour = volume of $H_2$ used (since 2 volumes $H_2$ → 2 volumes $H_2O$).

Mistake 4: Incorrectly Applying the Volume Ratio

The mistake:

Students think: “10 volumes $H_2$ + 5 volumes $O_2$ = 15 volumes product” — adding volumes directly.

Why it’s wrong:

Volumes are not additive in a chemical reaction. The product volume is determined by the coefficient ratio, not the sum of reactants.

How to avoid: …