JEE Mains Chemistry · Some Basic Concepts of Chemistry

Reaction Stoichiometry and Limiting Reagent

Mass, mole and volume relations from a balanced equation, with purity and yield, and finding which reactant runs out first.

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Why this matters

Thirty-four PYQs, half of them multiple choice and six from 2026, which makes this the chapter's largest page. Eighteen turn a mass of one substance into a mass, mole or volume of another, often with a purity or a yield. Sixteen give two reactants and ask which runs out first and how much of the other is left. Both rest on the same mole ratio.

Concept 1 of 2: Mole ratios, purity and yield

A balanced equation is a recipe in moles. Convert the given mass to moles, scale by the ratio of coefficients, and convert back. Purity trims the amount you start with; yield trims the amount you end with.

Definition

For $a\,\mathrm{A}\rightarrow b\,\mathrm{B}$ : $n_B=n_A\times\frac{b}{a}$ . Balance the equation first.
Purity: use only the pure part, $m_{\text{pure}}=\frac{\%}{100}\times m$ .
$\%\text{ yield}=\frac{\text{actual}}{\text{theoretical}}\times100$ .
Residues on heating: $\mathrm{CaCO_3\rightarrow CaO}$ keeps 56 g of every 100 g; $\mathrm{MgCO_3\rightarrow MgO}$ keeps 40 g of every 84 g.
$\mathrm{Ag_2CO_3}$ leaves metallic silver, because the $\mathrm{Ag_2O}$ formed also decomposes.
A mixture of two carbonates: two unknowns, two equations (total mass and residue mass).

Mass to mass

m_B=\frac{m_A}{M_A}\times\frac{b}{a}\times M_B

Worked example

What mass of

\mathrm{O_2}

burns

11

g of propane, and what mass of

\mathrm{CO_2}

forms?

\mathrm{C_3H_8+5O_2\rightarrow3CO_2+4H_2O}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q26Moderate

Example 1 · Some Basic Concepts of Chemistry · Reaction Stoichiometry and Limiting Reagent

The mass of iron converted into

Fe_{3}O_{4}

by the action of 18 g of steam is : (Given : Molar mass of H, O and Fe are 1, 16 and

56\text{ }g{\text{ }mol}^{- 1}

respectively) Assume iron is present in excess :

Silver carbonate leaves silver

\mathrm{Ag_2CO_3}

gives

\mathrm{Ag_2O}

, and

\mathrm{Ag_2O}

loses its oxygen too on strong heating. The residue is metallic silver: 2 mol of Ag per mole of carbonate.

Using the impure mass

A '75% pure' sample reacts only through its pure part. Scale the mass down before converting to moles, not after.

Drill 17 more on mole ratios, purity and yield

Concept 2 of 2: Limiting reagent and the excess left

Divide each reactant's moles by its coefficient. The smallest result runs out first and fixes how much product forms. The other reactant is left over by whatever it did not need.

Definition

Limiting reagent: the smallest $\frac{n}{\text{coefficient}}$ , not the smallest $n$ .
Work out every product from the limiting reagent only.
Excess left $=$ supplied $-$ used, where used $=$ the limiting moles $\times$ the coefficient ratio.
The molar mass of $\mathrm{AB_2}$ is $M_A+2M_B$ .
Gases in a closed vessel at fixed volume and temperature: partial pressures behave like moles.

Limiting reagent test

\text{limiting reagent}=\text{the smallest }\frac{n_i}{\nu_i}

Worked example

4.8

g of Mg is heated with

4.0

g of

\mathrm{O_2}

\mathrm{2Mg+O_2\rightarrow2MgO}

. Find the limiting reagent, the mass of MgO and the

\mathrm{O_2}

left.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q28Moderate

Example 2 · Some Basic Concepts of Chemistry · Reaction Stoichiometry and Limiting Reagent

A + 2\text{ }B \longrightarrow AB_{2}

36.0 g of 'A' (Molar mass :

60\text{ }g{\text{ }mol}^{- 1}

) and 56.0 g of ' B ' (Molar mass :

80\text{ }g{\text{ }mol}^{- 1}

) are allowed to react. Which of the following statements are correct? (A) ' A ' is the limiting reagent (B) 77.0 g of

{AB}_{2}

is formed (C) Molar mass of

{AB}_{2}

140\text{ }g{\text{ }mol}^{- 1}

(D) 15.0 g of A is left unreacted after the completion of reaction. Choose the correct answer from the options given below :

Fewest moles is not the test

Compare

\frac{n}{\text{coefficient}}

. In

\mathrm{N_2+3H_2}

0.5

mol of

\mathrm{N_2}

and

1.2

mol of

\mathrm{H_2}

\mathrm{H_2}

runs out (

0.4<0.5

) although it has more moles.

Product from the excess reagent

Working the product from the reactant in excess gives a larger answer, and it is always one of the options.

Drill 15 more on limiting reagent and the excess left

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Mole ratios, purity and yield
Mass to mass
$m_B=\frac{m_A}{M_A}\times\frac{b}{a}\times M_B$
Limiting reagent and the excess left
Limiting reagent test
$\text{limiting reagent}=\text{the smallest }\frac{n_i}{\nu_i}$

Watch out for (4)

Silver carbonate leaves silver
→ Mole ratios, purity and yield
Using the impure mass
→ Mole ratios, purity and yield
Fewest moles is not the test
→ Limiting reagent and the excess left
Product from the excess reagent
→ Limiting reagent and the excess left

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