Stoichiometry — Part 2: Calculations from equations: visual lesson, ICSE Class 10 Chemistry (Science Paper 2)

Lesson

Reading an equation in moles

Part 2 — a balanced equation is a recipe. Its coefficients tell us the ratio in which moles react and form, and from moles we can reach masses, gas volumes and numbers of particles.

In this lesson you will

Read a balanced equation in moles, masses and gas volumes
Calculate the mass of a product from the mass of a reactant
Calculate the volume of a gas at STP from a mass, and a mass from a volume
Solve volume–volume and multi-step problems

Table reading the equation for hydrogen burning in oxygen as molecules, moles, masses and gas volumes
Read as	2H₂	O₂	2H₂O
Molecules	2	1	2
Moles	2 mol	1 mol	2 mol
Masses (H = 1, O = 16)	4 g	32 g	36 g
Gas volumes at STP	44.8 dm³	22.4 dm³	(liquid)

Concept

What a balanced equation tells us

For a balanced equation:

The coefficients give the mole ratio (and, for gases at the same temperature and pressure, the volume ratio).
Masses come from moles × molar mass; the total mass of reactants equals the total mass of products.
Gas volumes at STP come from moles × 22.4 dm³.

The coefficients are not a mass ratio: in 2H₂ + O₂ → 2H₂O, 4 g of hydrogen reacts with 32 g of oxygen.

Concept

The three-step method

Given → moles: moles of A = mass ÷ molar mass (or volume at STP ÷ 22.4).
Mole ratio: moles of B = moles of A × (coefficient of B ÷ coefficient of A).
Moles → wanted: mass of B = moles × molar mass; volume of B at STP = moles × 22.4; particles = moles × Avogadro’s number.

Always start from a balanced equation.

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