MHT-CET Chemistry · Some Basic Concepts of Chemistry

Stoichiometry and Concentration

A balanced equation is a recipe in moles: convert the given mass or gas volume to moles, scale by the coefficient ratio, and convert to the target — then read solution strength (H2O2 volume strength, % by mass) off the same mole bridge.

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Why this matters

Seven PYQs, and the workhorse is one skill: mass-or-volume to moles, scale by the balanced-equation ratio, convert to what is asked. The KClO3 to O2, Mg + HCl and burning-carbon mass problems (four of the seven) are all this one move. Two questions test gas reactions where the volumes are already in the coefficient ratio (CH4 combustion, N2 + 3H2), which brings in limiting-reagent thinking. One HARD question turns H2O2 volume strength into % by mass — a two-formula chain that CET repeats almost every year.

Concept 1 of 3: Mole ratios from a balanced equation

The big coefficients in a balanced equation are a ratio of moles, never of grams. So every problem is the same three-step bridge: turn the known mass or gas volume into moles, scale by the coefficient ratio, then turn the target moles back into grams (or molecules).

Definition

Stoichiometry in three steps:

Balance the equation; the coefficients are the mole ratio of reactants to products.
Convert the known quantity to moles: mass gives $n = m/M$ ; a gas volume at STP gives $n = V/22.4$ (with $V$ in $\text{dm}^3$ ).
Scale by the ratio, then convert the target moles to what is asked — mass $= n\times M$ , or molecules $= n\times 6.022\times 10^{23}$ .
For $2\text{KClO}_3 \to 2\text{KCl} + 3\text{O}_2$ , the ratio $\text{KClO}_3 : \text{O}_2$ is 2 : 3, so $n_{\text{KClO}_3} = \tfrac{2}{3}\,n_{\text{O}_2}$ .

Mass of product from a known reactant quantity

m_{\text{target}} = n_{\text{known}} \times \dfrac{\text{coeff}_{\text{target}}}{\text{coeff}_{\text{known}}} \times M_{\text{target}}

$n_{\text{known}}$ moles of the given substance (mass/M or volume/22.4)
$\text{coeff}$ the balanced-equation coefficient of each substance
$M_{\text{target}}$ molar mass of the substance being found

Worked example

What mass of aluminium reacts with excess HCl to liberate 3.36 dm3 of hydrogen at STP? 2Al + 6HCl -> 2AlCl3 + 3H2 (Al = 27 g/mol).

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q97Moderate

Example 1 · Some Basic Concepts of Chemistry · Stoichiometry and Concentration

Find the mass of potassium chlorate required to liberate

5.6dm^{3}

of oxygen gas at STP? (molar mass of

KClO_{3}= 122.5\text{ }g/mol

)

Coefficients are moles, not grams

The 2 : 3 in

2\text{KClO}_3 \to 3\text{O}_2

is a mole ratio. You cannot put masses straight into it — always convert to moles first, scale, then convert back. A gram-to-gram shortcut only works by accident when the molar masses happen to match.

Do not skip the fractional ratio

For

0.25

mol

\text{O}_2

, the moles of

\text{KClO}_3

are

\tfrac{2}{3}\times 0.25

, not

0.25

. Dropping the

\tfrac{2}{3}

gives

0.25\times 122.5 \approx 30.6

g — a wrong distractor. Keep the ratio the right way up: fewer moles of

\text{KClO}_3

than of

\text{O}_2

Drill 3 more on mole ratios from a balanced equation

Concept 2 of 3: Combining gaseous volumes and the limiting reagent

For gases measured at the same temperature and pressure, volumes behave exactly like moles — so you can skip the mole conversion and work directly in the coefficient ratio (Gay-Lussac's law of combining volumes, which follows from Avogadro's law). The one extra check is which reactant runs out first: the limiting reagent decides how much product forms.

Definition

Gas-phase stoichiometry shortcut (same T and P):

By Avogadro's law, equal volumes hold equal numbers of molecules, so gas volumes are in the same ratio as the balanced-equation coefficients.
For $\text{CH}_4 + 2\text{O}_2 \to \text{CO}_2 + 2\text{H}_2\text{O}$ , each volume of $\text{CH}_4$ needs 2 volumes of $\text{O}_2$ .
Limiting reagent: divide each reactant's supplied amount by its coefficient; the smallest quotient is the reagent that runs out and controls the product.
Product volume $=$ (limiting reactant volume) $\times$ (product coefficient / limiting-reactant coefficient).

Product volume from a limiting gaseous reactant

V_{\text{product}} = V_{\text{limiting}} \times \dfrac{\text{coeff}_{\text{product}}}{\text{coeff}_{\text{limiting}}}

Worked example

What volume of ammonia forms when 10 dm3 of N2 reacts with 30 dm3 of H2 at the same temperature and pressure? N2 + 3H2 -> 2NH3.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q85Moderate

Example 2 · Some Basic Concepts of Chemistry · Stoichiometry and Concentration

What volume of ammonia is formed when 10 dm³ dinitrogen reacts with 30 dm³ dihydrogen at same temperature and pressure?

Identify the limiting reagent before scaling

Do not scale off the reactant you happen to see first. Divide each supplied volume by its coefficient and use the smallest quotient. In

\text{N}_2 + 3\text{H}_2

, 30 dm

^3

\text{H}_2

only reacts with 10 dm

^3

\text{N}_2

; if you had 40 dm

^3

\text{H}_2

\text{N}_2

would still cap the product at 20 dm

^3

\text{NH}_3

Volumes need the same T and P

The volume-equals-mole shortcut only holds when all gases are at the same temperature and pressure. If a question gives one gas at STP and asks about another under different conditions, convert to moles first.

Drill 1 more on combining gaseous volumes and the limiting reagent

Concept 3 of 3: Concentration: percent by mass and H2O2 volume strength

Percent by mass is just the mass of solute per 100 g of solution. Hydrogen peroxide is sold by 'volume strength' — the litres of O2 one litre of the solution releases on decomposition — and there is a fixed conversion from that number to molarity, from which % by mass follows.

Definition

Two linked results the CET repeats:

Percent by mass $= \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$ .
Volume strength of $\text{H}_2\text{O}_2$ : an 'X volume' solution releases X litres of $\text{O}_2$ per litre of solution at STP.
Because $2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2$ , this fixes molarity as $M = \dfrac{\text{volume strength}}{11.2}$ .
Then in 1 L (mass $1000$ g at density $1$ g/mL): mass of $\text{H}_2\text{O}_2 = M\times 34$ , so % by mass $= \dfrac{M\times 34}{1000}\times 100$ .

H2O2 molarity from volume strength, and percent by mass

M_{\text{H}_2\text{O}_2} = \dfrac{\text{volume strength}}{11.2} \qquad \%\text{ by mass} = \dfrac{\text{mass of solute}}{\text{mass of solution}}\times 100

Worked example

Calculate the percent by mass of an H2O2 solution that is '67.2 volume'. (H2O2 = 34 g/mol, solution density = 1 g/mL.)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q88Hard

Example 3 · Some Basic Concepts of Chemistry · Stoichiometry and Concentration

Calculate

\%

by mass of a

H_{2}O_{2}

solution that is 67.2 by volume.

Divide volume strength by 11.2, not 22.4

One mole of

\text{H}_2\text{O}_2

releases only half a mole of

\text{O}_2

(

2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2

), so 1 mol/L gives

11.2

L of

\text{O}_2

per litre — hence

M = \text{volume strength}/11.2

. Using

22.4

halves your answer.

Percent by mass uses the mass of solution, not solvent

The denominator is the total mass of the solution (solute + solvent), which is why 1 L at density 1 g/mL is taken as 1000 g. Dividing by the mass of water alone gives a wrong, slightly larger percentage.

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 (3)

Mole ratios from a balanced equation
Mass of product from a known reactant quantity
$m_{\text{target}} = n_{\text{known}} \times \dfrac{\text{coeff}_{\text{target}}}{\text{coeff}_{\text{known}}} \times M_{\text{target}}$
Combining gaseous volumes and the limiting reagent
Product volume from a limiting gaseous reactant
$V_{\text{product}} = V_{\text{limiting}} \times \dfrac{\text{coeff}_{\text{product}}}{\text{coeff}_{\text{limiting}}}$
Concentration: percent by mass and H2O2 volume strength
H2O2 molarity from volume strength, and percent by mass
$M_{\text{H}_2\text{O}_2} = \dfrac{\text{volume strength}}{11.2} \qquad \%\text{ by mass} = \dfrac{\text{mass of solute}}{\text{mass of solution}}\times 100$

Watch out for (6)

Coefficients are moles, not grams
→ Mole ratios from a balanced equation
Do not skip the fractional ratio
→ Mole ratios from a balanced equation
Identify the limiting reagent before scaling
→ Combining gaseous volumes and the limiting reagent
Volumes need the same T and P
→ Combining gaseous volumes and the limiting reagent
Divide volume strength by 11.2, not 22.4
→ Concentration: percent by mass and H2O2 volume strength
Percent by mass uses the mass of solution, not solvent
→ Concentration: percent by mass and H2O2 volume strength

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Related notes

NDA Chemistry: Stoichiometry and the Laws of Chemical Combination →