Q.Assertion (A): Combustion of 16 g of methane gives 18 g of water.
Reason (R): In the combustion of methane, water is one of the products.
Choose the correct option out of the choices given below.

Concept: Stoichiometry of combustion reactions and molar mass calculations.

The combustion of methane follows:

One mole of methane (16 g) produces two moles of water. The molar mass of water is 18 g/mol, so two moles yield $2 \times 18 = 36$ g of water, not 18 g.

Assertion (A) is false — combustion of 16 g methane produces 36 g water.

Reason (R) is true — water is indeed a product of methane combustion. …

Understanding the Problem

This question tests two things: your ability to perform stoichiometric calculations and your understanding of how assertion-reason pairs work. The reason statement is obviously true—methane combustion does produce water—but that doesn't automatically make the assertion correct. We need to verify the mass claim independently.

The balanced equation for methane combustion is:

This tells us the molar ratio: one mole of methane produces two moles of water. Now we convert the given mass to moles, apply the stoichiometry, and check whether 18 g of water is the correct answer.

Step-by-Step Verification

1. Calculate moles of methane burned

The molecular mass of methane $\ce{CH4}$ is:

Given mass = 16 g, so:

2. Apply stoichiometry to find moles of water produced

From the balanced equation, 1 mol $\ce{CH4}$ produces 2 mol $\ce{H2O}$:

3. Convert moles of water to mass

The molecular mass of water $\ce{H2O}$ is:

Mass of water produced:

$m_{\ce{H2O}} = 2 \times 18 = 36 \text{ g}$ …

Concept: Stoichiometry of Combustion Reactions

The relevant concept is stoichiometric calculation based on a balanced chemical equation — specifically, using mole ratios to predict the mass of products from a given mass of reactant.

Method: Mole-to-Mass Stoichiometry

Steps:

1. Write and balance the chemical equation for combustion of methane:

2. Find molar masses:

   • Methane ($\text{CH}_4$): $12 + 4 \times 1 = 16 \text{ g/mol}$
   • Water ($\text{H}_2\text{O}$): $2 \times 1 + 16 = 18 \text{ g/mol}$

3. Convert given mass of methane to moles:

4. Use mole ratio from balanced equation:

   • From equation: $1 \text{ mol CH}_4$ produces $2 \text{ mol H}_2\text{O}$
   • So, $1 \text{ mol CH}_4$ gives $2 \text{ mol H}_2\text{O}$

5. Convert moles of water to mass: …

🔍 The Concept: Stoichiometry of Combustion

The balanced equation for combustion of methane is:

From this:

• 1 mole of $\text{CH}_4$ (16 g) produces 2 moles of $\text{H}_2\text{O}$ (2 × 18 g = 36 g).

So the Assertion (A) — that 16 g of methane gives 18 g of water — is false.

The Reason (R) — that water is a product — is true, but it does not correctly explain the false assertion.

Correct option: A is false, R is true.

✗ Common Mistakes Students Make

1. Forgetting the mole ratio

• Mistake: Assuming 1 mole of reactant gives 1 mole of product.
• Why it happens: Students see “CH₄ → H₂O” and think 1:1.
• How to avoid: Always write the balanced chemical equation first. Count coefficients carefully.

2. Confusing mass with moles

• Mistake: Thinking 16 g CH₄ gives 18 g H₂O because 16 + something = 18.
• Why it happens: Misapplying the law of conservation of mass without considering the other product (CO₂).
• How to avoid: Remember — mass is conserved overall, but each product’s mass depends on the mole ratio. Use:

3. Not checking the Assertion-Reason logic

• Mistake: Choosing “Both A and R are true, and R is the correct explanation of A” just because R sounds correct.
• Why it happens: Students focus on whether each statement is individually true, ignoring the link between them.
• How to avoid: First check truth of A and R separately. Then check if R correctly explains A. If A is false, R cannot explain it.

4. Miscalculating molar masses …