The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2A + 4B \rightarrow 3C + 4D$, when 5 moles of A react with 6 moles of B, then (i) which is the limiting reagent? (ii) calculate the amount of C formed?

The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A + 4B arro

NCERT Exemplar · Q35

2A + 4B \rightarrow 3C + 4D

(i) which is the limiting reagent?

(ii) calculate the amount of C formed?

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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: …

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The limiting reagent is the reactant that runs out first, stopping the reaction. Here B is limiting because it requires 10 moles to fully react with 5 moles of A but only 6 are available; 4.5 moles of C form.

Why limiting reagents matter

A balanced equation tells you the ratio in which reactants combine, not the amounts you must use. In practice you rarely have reactants in exactly the stoichiometric ratio. One reactant will be completely consumed first—that's the limiting reagent—and it determines how much product forms. The other reactant(s) remain in excess.

The strategy is simple: for each reactant, calculate how much product it alone could make if it were the only constraint. Whichever gives the smallest amount of product is limiting.

Step-by-step solution

1. Write down what the stoichiometry demands

The balanced equation is

2A + 4B \rightarrow 3C + 4D

This tells us:

2 moles of $A$ require 4 moles of $B$ (a 1:2 ratio)
2 moles of $A$ produce 3 moles of $C$
4 moles of $B$ produce 3 moles of $C$

2. Check how much B is needed to react with all the A

We have 5 moles of $A$ . According to the stoichiometry, 2 moles of $A$ need 4 moles of $B$ , so:

\text{Moles of } B \text{ required} = 5 \times \frac{4}{2} = 10 \text{ moles}

But we only have 6 moles of $B$ available. Not enough $B$ to consume all the $A$ .

3. Check how much A is needed to react with all the B

We have 6 moles of $B$ . According to the stoichiometry, 4 moles of $B$ need 2 moles of $A$ , so:

\text{Moles of } A \text{ required} = 6 \times \frac{2}{4} = 3 \text{ moles}

We have 5 moles of $A$ available, which is more than enough.

Watch out

A common mistake is to compare the absolute amounts (5 vs. 6) and conclude that $A$ is limiting because 5 < 6. You must compare amounts relative to the stoichiometric ratio.

4. Identify the limiting reagent …

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✗ Mistake 1: Comparing moles of reactants directly without considering stoichiometric coefficients

What students do wrong:

They see 5 moles of A and 6 moles of B, and think “B has more moles, so A is limiting” — or vice versa.

Why it’s wrong:

The reaction requires 2 moles of A for every 4 moles of B. You cannot compare raw mole numbers — you must compare the mole ratio to the balanced equation.

How to avoid:

Always calculate the required moles of one reactant based on the other.

From the equation: $2A + 4B \rightarrow 3C + 4D$
For 5 moles of A, required B = $5 \times \frac{4}{2} = 10$ moles
But we only have 6 moles of B — so B is limiting.

✓ Key rule:

Limiting reagent is the one that runs out first when you check the stoichiometric ratio.

✗ Mistake 2: Using the wrong reactant’s moles to calculate product formed

What students do wrong:

After identifying the limiting reagent, they sometimes use the excess reactant to calculate the amount of product.

Why it’s wrong:

The product amount is determined only by the limiting reagent — the excess reactant doesn’t fully react.

How to avoid:

Once you identify the limiting reagent (here, B), use only its moles to find moles of C.

From equation: $4$ moles of B produce $3$ moles of C
So, $6$ moles of B produce:

\text{moles of C} = 6 \times \frac{3}{4} = 4.5 \text{ moles}

✓ Key rule:

Product amount = (moles of limiting reagent) × (stoichiometric coefficient of product / coefficient of limiting reagent)

✗ Mistake 3: Forgetting to convert to mass when asked for “amount”

What students do wrong:

They stop at moles of C, but the question asks for “amount” — which in many exam contexts means mass in grams.

Why it’s wrong:

“Amount” can be ambiguous, but in numerical problems, it usually means mass unless specified otherwise.

How to avoid:

Always check the question. If it says “amount of C formed”, and molar mass is given (or can be assumed from atomic masses), convert:

Molar mass of C (say, from atomic masses: e.g., C = 12 g/mol)
Mass of C = $4.5 \times 12 = 54$ g

If no molar mass is given, state the answer in moles and mention that mass requires molar mass. …

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Q.The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2A + 4B \rightarrow 3C + 4D$ , when 5 moles of A react with 6 moles of B, then

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: …

Why limiting reagents matter

Step-by-step solution

Concept: Limiting Reagent

Method: Mole Ratio Comparison Method

Steps:

✗ Mistake 1: Comparing moles of reactants directly without considering stoichiometric coefficients

✗ Mistake 2: Using the wrong reactant’s moles to calculate product formed

✗ Mistake 3: Forgetting to convert to mass when asked for “amount”

Q.The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A+4B→3C+4D2A + 4B \rightarrow 3C + 4D2A+4B→3C+4D, when 5 moles of A react with 6 moles of B, then

Why limiting reagents matter

Step-by-step solution

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Q.The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2A + 4B \rightarrow 3C + 4D$ , when 5 moles of A react with 6 moles of B, then

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