Which of the following statements is correct about the reaction given below: 4Fe(s) + 3O₂(g) arrow 2Fe₂O₃(g) (

NCERT Exemplar · Q13

4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(g)

(i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass.

(ii) Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed.

(iii) Amount of

Fe_2O_3

can be increased by taking any one of the reactants (iron or oxygen) in excess.

(iv) Amount of

Fe_2O_3

produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

Telangana TsbieMCQ· 1mImportance★★★★★est

64% · 57/89 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

Imagine you're at a market buying apples. You don't weigh each apple individually — you put a dozen on a scale. The total weight tells you something about the apples, but it also depends on how many apples you have.

Atoms and molecules are unimaginably tiny. A single water molecule ( $H_2O$ ) weighs about $3 \times 10^{-23}$ grams. That number is useless for practical chemistry. So instead of working with individual molecules, chemists count them in huge fixed numbers — specifically, $6.022 \times 10^{23}$ molecules, which is called one mole.

Molecular mass is simply the mass of one mole of a substance, expressed in grams per mole (g/mol). It answers the question: "If I have $6.022 \times 10^{23}$ molecules of this compound, how much would they weigh on a lab balance?"

Note

The number $6.022 \times 10^{23}$ is Avogadro's constant. It's chosen so that the mass of one mole of carbon-12 atoms is exactly 12 grams — matching the atomic mass unit scale perfectly.

The Precise Definition

Molecular mass (also called molar mass) is the mass of one mole of a molecular substance. It is numerically equal to the sum of the atomic masses of all atoms in the molecule, expressed in g/mol.

For example:

Water ( $H_2O$ ): 2 hydrogen atoms + 1 oxygen atom
Atomic mass of H = 1.008 g/mol
Atomic mass of O = 16.00 g/mol
Molecular mass of $H_2O$ = $2(1.008) + 16.00 = 18.016$ g/mol

\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass of that element})

How to Calculate It — Step by Step

Let's take glucose, $C_6H_{12}O_6$ , as a worked example.

Step 1: Identify each element and its count

Carbon (C): 6 atoms
Hydrogen (H): 12 atoms
Oxygen (O): 6 atoms

Step 2: Look up atomic masses (from the periodic table)

C: 12.01 g/mol
H: 1.008 g/mol
O: 16.00 g/mol

Step 3: Multiply and add

\text{Molecular mass} = 6(12.01) + 12(1.008) + 6(16.00)

= 72.06 + 12.096 + 96.00

= 180.156 \text{ g/mol}

Tip

Always keep at least 2 decimal places from the periodic table. For exam problems, they usually give you atomic masses — use exactly what's provided.

Why This Matters

Molecular mass is the bridge between the microscopic world (atoms and molecules) and the macroscopic world (grams you can weigh). Once you know the molecular mass, you can:

Convert grams to moles: $\text{moles} = \frac{\text{mass in grams}}{\text{molecular mass}}$
Convert moles to grams: $\text{mass} = \text{moles} \times \text{molecular mass}$
Determine the number of molecules: $\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

Watch out

Do not confuse molecular mass with atomic mass. Atomic mass refers to a single element (like oxygen = 16.00 g/mol). Molecular mass refers to a compound (like $CO_2$ = 44.01 g/mol). Also, for ionic compounds like NaCl, we use formula mass (same calculation, but the substance isn't molecular).

Common Exam Pitfalls

Forgetting to multiply by the subscript. In $H_2SO_4$ , there are 2 hydrogens, not 1. …

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.

1. The Core Question: What is a Mole?

A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains $6.022 \times 10^{23}$ particles (Avogadro's number, $N_A$ ).

Why this number?

It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.

Example: 1 atom of carbon-12 has mass 12 amu.
1 mole of carbon-12 has mass 12 grams.

This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).

2. The Fundamental Relationship

The key formula is:

n = \frac{m}{M}

Where:

$n$ = number of moles
$m$ = mass of substance (in grams)
$M$ = molar mass (in g/mol)

Why does this work?

Think of it as a conversion factor:

If 1 mole of a substance weighs $M$ grams, then $m$ grams contains $\frac{m}{M}$ moles.

Derivation logic:

Molar mass $M$ tells you: "1 mol = $M$ g"
So the conversion factor is $\frac{1 \text{ mol}}{M \text{ g}}$
Multiply mass $m$ by this factor: $m \times \frac{1}{M} = \frac{m}{M}$ moles

3. Connecting to Number of Particles

n = \frac{N}{N_A}

Where:

$N$ = number of particles (atoms, molecules, ions)
$N_A = 6.022 \times 10^{23}$ particles/mol

Why?

1 mole = $N_A$ particles
So $N$ particles = $\frac{N}{N_A}$ moles

Combine both formulas:

\frac{m}{M} = \frac{N}{N_A}

This single equation ties mass, molar mass, number of particles, and Avogadro's number together.

4. The Gas Volume Connection (for gases at STP)

For gases only:

n = \frac{V}{22.4 \text{ L/mol}}

Why 22.4 L?

From the ideal gas law: $PV = nRT$

At STP (Standard Temperature and Pressure: 0°C, 1 atm):

$P = 1$ atm
$T = 273.15$ K
$R = 0.0821$ L·atm/(mol·K)

For $n = 1$ mole:

V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273.15}{1} \approx 22.4 \text{ L}

So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.

5. The Stoichiometry Chain: From One Substance to Another

In a balanced chemical equation like:

aA + bB \rightarrow cC + dD

The coefficients tell you the mole ratio:

\frac{\text{moles of A}}{\text{moles of B}} = \frac{a}{b}

Why this works: …

Start your 14-day free trial to unlock the full solution →

The given balanced chemical reaction adheres to the Law of Conservation of Mass, meaning the total mass of reactants equals the total mass of products. Therefore, statement (i) is correct.

Understanding the Core Concept: Law of Conservation of Mass

At the heart of any chemical reaction lies the Law of Conservation of Mass. This fundamental law, proposed by Antoine Lavoisier, states that mass is neither created nor destroyed in any chemical reaction. This means that the total mass of all the reactants consumed in a chemical reaction must be exactly equal to the total mass of all the products formed.

A balanced chemical equation is a direct representation of this law. The coefficients in a balanced equation ensure that the number of atoms of each element on the reactant side is equal to the number of atoms of that same element on the product side. Since atoms are conserved, their total mass must also be conserved.

Let's analyze the given reaction and the provided statements based on this and other relevant chemical principles.

The reaction is:

$4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(g)$

To evaluate the statements, we will use the approximate atomic masses commonly used in calculations:

Atomic mass of Iron (Fe) $\approx 56 \text{ g/mol}$
Atomic mass of Oxygen (O) $\approx 16 \text{ g/mol}$

Step-by-Step Analysis of the Options

Evaluate Option (i): "Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass."

To check this, we calculate the total mass of reactants and products based on the stoichiometry of the balanced equation.
- Mass of Reactants:
  - Mass of $4Fe$ : $4 \times (\text{atomic mass of Fe}) = 4 \times 56 \text{ g/mol} = 224 \text{ g}$
  - Mass of $3O_2$ : $3 \times (\text{molecular mass of } O_2) = 3 \times (2 \times 16 \text{ g/mol}) = 3 \times 32 \text{ g/mol} = 96 \text{ g}$
  - Total mass of reactants = $224 \text{ g} + 96 \text{ g} = 320 \text{ g}$
- Mass of Products:
  - Mass of $2Fe_2O_3$ : $2 \times (\text{molecular mass of } Fe_2O_3)$
  - Molecular mass of $Fe_2O_3 = (2 \times \text{atomic mass of Fe}) + (3 \times \text{atomic mass of O})$
  - Molecular mass of $Fe_2O_3 = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \text{ g/mol}$
  - Total mass of products = $2 \times 160 \text{ g/mol} = 320 \text{ g}$
Since the total mass of reactants ( $320 \text{ g}$ ) equals the total mass of products ( $320 \text{ g}$ ), the reaction indeed follows the Law of Conservation of Mass. Therefore, statement (i) is correct.

Law of Conservation of Mass:

$\text{Total mass of reactants} = \text{Total mass of products}$
Evaluate Option (ii): "Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed."

The first part of the statement, "Total mass of reactants = total mass of product," is correct, as established in step 1. However, the conclusion, "therefore, law of multiple proportions is followed," is incorrect.

The Law of Multiple Proportions applies when two elements combine to form more than one compound. It states that if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers.

In the given reaction, iron and oxygen combine to form only one product, $Fe_2O_3$ . This reaction itself does not demonstrate the Law of Multiple Proportions. While the Law of Conservation of Mass is always followed in a chemical reaction, it does not imply the Law of Multiple Proportions.
Evaluate Option (iii): "Amount of $Fe_2O_3$ can be increased by taking any one of the reactants (iron or oxygen) in excess."

This statement deals with the concept of limiting and excess reactants. In a chemical reaction, the limiting reactant is the one that is completely consumed first, thereby determining the maximum amount of product that can be formed. The excess reactant is the one present in a greater amount than required to react with the limiting reactant. …

Start your 14-day free trial to unlock the full solution →

Here’s a breakdown of the common mistakes students make with this question and how to avoid each.

Mistake 1: Confusing the Law of Conservation of Mass with the Law of Multiple Proportions

What students do wrong:

They see “total mass of reactants = total mass of product” and immediately tick option (ii), thinking it’s the law of multiple proportions.

Why it’s wrong:

The law of multiple proportions deals with the ratios of masses of one element that combine with a fixed mass of another element to form different compounds. It does not say “total mass of reactants = total mass of product.” That statement is actually the law of conservation of mass.

How to avoid:

Memorise the exact wording of each law:
- Conservation of mass: Mass is neither created nor destroyed in a chemical reaction → total mass of reactants = total mass of products.
- Multiple proportions: When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.
In this reaction, only one product ( $Fe_2O_3$ ) is formed, so the law of multiple proportions does not apply at all.

Correct approach for option (i):

Check if the equation is balanced.

Left: 4 Fe, 6 O
Right: 4 Fe, 6 O Yes, it’s balanced → mass is conserved. So (i) is correct.

Mistake 2: Thinking “excess reactant” always increases product

What students do wrong:

They assume that adding more of any reactant will increase the amount of product. So they pick option (iii).

Why it’s wrong:

In a chemical reaction, the limiting reactant determines how much product is formed. If you already have enough of one reactant, adding more of it will not increase the product — the other reactant becomes the limiting factor.

How to avoid:

Remember: Only the limiting reactant limits the product.
Adding excess of the reactant that is already in excess does nothing to the yield.
In this reaction, if oxygen is in excess, adding more oxygen won’t increase $Fe_2O_3$ — iron is the limiting reactant.

Correct approach for option (iii):

It says “by taking any one of the reactants in excess.” This is false because if you take the wrong one (the one already in excess), product doesn’t increase. So (iii) is incorrect.

Mistake 3: Believing excess reactant decreases product

What students do wrong:

They think adding extra reactant will “use up” the other reactant or somehow reduce the yield, so they pick option (iv).

Why it’s wrong: …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
$Q. An organic compound on analysis is found to have 10.06% carbon, 0.84% hydrogen and 89.10% chlorine by weight. The simplest whole number ratio of C, H and Cl is (A) 1:2:3 (B) 1:1:3 (C) 1:2:2 (D) 1:3:1$
›Reveal solutionSolution
The problem asks for the simplest whole‑number ratio of C, H, and Cl from given weight percentages. By converting percentages to moles and dividing by the smallest mole count, we obtain the ratio 1 : 1 : 3, which corresponds to option (B).

Concept & Intuition

When we are given the percentage by weight of each element in a compound, the “simplest whole‑number ratio” is found by converting those masses into moles. Why? Because chemical formulas count atoms, not grams. The mole is the bridge between mass and number of atoms. Once we have the mole amounts, we divide by the smallest to get the smallest integer ratio.

Step‑by‑step reasoning

Assume a 100 g sample – Percentages become grams directly.

Carbon: $10.06\%$ → $10.06\ \text{g}$

Hydrogen: $0.84\%$ → $0.84\ \text{g}$

Chlorine: $89.10\%$ → $89.10\ \text{g}$

Convert each mass to moles using atomic masses (C = 12.01, H = 1.008, Cl = 35.45).

Moles of C: $\displaystyle \frac{10.06}{12.01} \approx 0.8376$

Moles of H: $\displaystyle \frac{0.84}{1.008} \approx 0.8333$

Moles of Cl: $\displaystyle \frac{89.10}{35.45} \approx 2.513$

Find the smallest mole value – Here it is hydrogen: $0.8333$ mol (very close to carbon’s 0.8376, but slightly smaller).

Divide each mole amount by the smallest to get a ratio:

C: $\displaystyle \frac{0.8376}{0.8333} \approx 1.005$ → essentially 1

H: $\displaystyle \frac{0.8333}{0.8333} = 1$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$_2$
›Reveal solutionSolution
The mole fraction of H₂SO₄ is 0.9, meaning 9 moles of acid per 1 mole of water. Converting to masses gives 882 g H₂SO₄ and 18 g water, so the mass percent is $\frac{882}{900} \times 100 = 98\%$ . The answer is (C).

The key here is to understand what mole fraction actually tells you. It’s a ratio of moles — not masses. So when you’re given a mole fraction of 0.9 for H₂SO₄ in water, it means that out of every 10 total moles in the solution, 9 are H₂SO₄ and 1 is H₂O. That’s the starting point.

Mass percent, on the other hand, is a ratio of masses. So you need to convert those moles into grams using the molar masses, then find what fraction of the total mass is acid.

Let’s walk through it.

Interpret the mole fraction.

Mole fraction of H₂SO₄, $x_{\text{H}_2\text{SO}_4} = 0.9$ .

This means $x_{\text{H}_2\text{O}} = 1 - 0.9 = 0.1$ .

The simplest way to work: assume a total of 1 mole of solution. Then:

Moles of H₂SO₄ = $0.9$ mol

Moles of H₂O = $0.1$ mol

(You could also scale it to 10 total moles — same result.)

Find the masses.

Molar mass of H₂SO₄:

$2 \times 1 + 32 + 4 \times 16 = 2 + 32 + 64 = 98$ g/mol

Mass of H₂SO₄ = $0.9 \times 98 = 88.2$ g

Molar mass of H₂O: $2 \times 1 + 16 = 18$ g/mol

Mass of H₂O = $0.1 \times 18 = 1.8$ g

Total mass of solution = $88.2 + 1.8 = 90.0$ g

Calculate mass percent. …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$X_3Y_5$
›Reveal solutionSolution

In an hcp lattice, the number of tetrahedral voids is twice the number of atoms. If Y occupies two-thirds of these voids, the ratio of Y to X is 4:3, giving the formula $X_3Y_4$ .

The key to this problem is understanding the geometry of a hexagonal close-packed (hcp) lattice and how tetrahedral voids relate to the number of atoms in the lattice. Many students memorise formulas without seeing why they work, so let’s build the reasoning from the ground up.

In any close-packed structure — whether hcp or ccp (fcc) — each sphere in the lattice touches its neighbours in a way that leaves gaps, or voids, between them. There are two types: octahedral voids and tetrahedral voids. For every atom in a close-packed lattice, there is exactly one octahedral void and two tetrahedral voids. This is a fixed geometric fact, not a coincidence — it comes from how the layers stack.

So if element X forms an hcp lattice, the number of X atoms is the number of lattice points. Let that number be $n$ . Then the number of tetrahedral voids available is $2n$ .

Now, element Y occupies two-thirds of these tetrahedral voids. That means:
$\text{Number of Y atoms} = \frac{2}{3} \times (2n) = \frac{4n}{3}$
We now have the ratio of Y to X:

$\frac{\text{Y}}{\text{X}} = \frac{4n/3}{n} = \frac{4}{3}$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
$A_xB_y$
›Reveal solutionSolution
In a hexagonal close-packed (hcp) lattice, there are 6 effective atoms per unit cell, and twice that number of tetrahedral voids. If element B forms the hcp lattice and element A occupies $2/3^{rd}$ of the tetrahedral voids, the compound's formula is $\boxed{A_4B_3}$ .

When elements combine to form crystalline solids, one type of atom often forms a regular lattice structure, and the other type of atom occupies the "empty spaces" or voids within that lattice. To determine the chemical formula of such a compound, we need to find the ratio of the number of atoms of each element present in the unit cell.

The key concepts here are:

Hexagonal Close-Packed (hcp) Lattice: This is a type of close-packed structure where atoms are arranged in a hexagonal pattern. In an hcp unit cell, the effective number of atoms is 6. These atoms form the basic framework of the crystal.

Voids in Close-Packed Structures: In any close-packed structure (like hcp or ccp/fcc), there are two main types of interstitial voids:

Octahedral voids: These are surrounded by 6 atoms. The number of octahedral voids is equal to the effective number of atoms in the lattice.

Tetrahedral voids: These are surrounded by 4 atoms. The number of tetrahedral voids is twice the effective number of atoms in the lattice.

In this problem, element B forms the hcp lattice, and element A occupies a fraction of the tetrahedral voids. By calculating the effective number of B atoms and then the number of A atoms based on the void occupation, we can establish their ratio and thus the compound's formula.

Here's how we determine the formula:

Determine the effective number of B atoms:

Element B forms the hcp lattice. For an hcp unit cell, the effective number of atoms is 6.

So, the number of B atoms per unit cell, $N_B = 6$ .

Calculate the total number of tetrahedral voids: …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
$1.24~\mathrm{g/L}$
›Reveal solutionSolution
To identify the gas, we use the ideal gas law to calculate its molar mass from the given density at standard temperature and pressure. The calculated molar mass is approximately $27.8~\mathrm{g/mol}$ , which corresponds to carbon monoxide ( $CO$ ).

The density of a gas is directly related to its molar mass, temperature, and pressure. This relationship is derived from the ideal gas law, which describes the behavior of most gases under typical conditions. By knowing the density of a gas at specific temperature and pressure, we can determine its molar mass and, consequently, its identity.

Here's how to approach this problem:

Understand the Ideal Gas Law and its relation to density.

The ideal gas law is given by $PV = nRT$ , where:

$P$ is pressure

$V$ is volume

$n$ is the number of moles

$R$ is the ideal gas constant

$T$ is temperature in Kelvin

We know that the number of moles ( $n$ ) can be expressed as the mass ( $m$ ) of the gas divided by its molar mass ( $M$ ): $n = \frac{m}{M}$ .

Substituting this into the ideal gas law gives:

$PV = \frac{m}{M}RT$

Rearranging this equation to solve for density ( $\rho = \frac{m}{V}$ ):

$P = \frac{m}{V} \frac{RT}{M}$

$P = \rho \frac{RT}{M}$

Finally, we can express density in terms of molar mass, pressure, and temperature:
$\rho = \frac{PM}{RT}$
Or, to find the molar mass:

$M = \frac{\rho RT}{P}$

Identify the given values and standard conditions.

The problem provides the following information:

Density ( $\rho$ ) $= 1.24~\mathrm{g/L}$

Temperature ( $T$ ) $= 0~^\circ\mathrm{C}$

Pressure ( $P$ ) $= 1~\mathrm{atm}$

These conditions ( $0~^\circ\mathrm{C}$ and $1~\mathrm{atm}$ ) are known as Standard Temperature and Pressure (STP).

Important

For calculations involving the ideal gas law, temperature must always be in Kelvin.

$T(\mathrm{K}) = T(^\circ\mathrm{C}) + 273.15$

So, $T = 0~^\circ\mathrm{C} + 273.15 = 273.15~\mathrm{K}$ .

We need to choose the appropriate value for the ideal gas constant ( $R$ ). Since pressure is in atmospheres ( $\mathrm{atm}$ ) and volume is implied in liters ( $\mathrm{L}$ ) from the density unit ( $\mathrm{g/L}$ ), we use:

$R = 0.0821~\mathrm{L \cdot atm / (mol \cdot K)}$

Calculate the molar mass ( $M$ ) of the gas.

Using the rearranged formula $M = \frac{\rho RT}{P}$ :

$M = \frac{(1.24~\mathrm{g/L}) \times (0.0821~\mathrm{L \cdot atm / (mol \cdot K)}) \times (273.15~\mathrm{K})}{(1~\mathrm{atm})}$

Let's perform the calculation:

$M = 1.24 \times 0.0821 \times 273.15~\mathrm{g/mol}$

$M \approx 27.79~\mathrm{g/mol}$ …
Start your 14-day free trial to see the full solution →

←Previous questionQ12 — Which of the following statements about a compound is incorrect? (i) A…→Next questionQ14 — Which of the following reactions is not correct according to the law o…

Q.Which of the following statements is correct about the reaction given below: $4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(g)$

Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

The Precise Definition

How to Calculate It — Step by Step

Why This Matters

Common Exam Pitfalls

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

1. The Core Question: What is a Mole?

2. The Fundamental Relationship

Why does this work?

3. Connecting to Number of Particles

Why?

4. The Gas Volume Connection (for gases at STP)

Why 22.4 L?

5. The Stoichiometry Chain: From One Substance to Another

Why this works: …

Understanding the Core Concept: Law of Conservation of Mass

Step-by-Step Analysis of the Options

Concept: Laws of Chemical Combination & Limiting Reagent

Method: Law of Conservation of Mass & Stoichiometric Analysis

Mistake 1: Confusing the Law of Conservation of Mass with the Law of Multiple Proportions

Mistake 2: Thinking “excess reactant” always increases product

Mistake 3: Believing excess reactant decreases product

Q.Which of the following statements is correct about the reaction given below: 4Fe(s)+3O2(g)→2Fe2O3(g)4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(g)4Fe(s)+3O2​(g)→2Fe2​O3​(g)

Understanding the Core Concept: Law of Conservation of Mass

Step-by-Step Analysis of the Options

🎓Unlock everything free for 14 days

Q.Which of the following statements is correct about the reaction given below: $4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(g)$

Unlock everything free for 14 days