Moles, reacting masses and limiting reactants (HT)

Chemistry · Quantitative chemistry. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Chemists count particles in moles. One mole of any substance contains 6.02 × 10²³ particles (the Avogadro constant). The mass of one mole in grams is equal to the relative formula mass, so one mole of water (M_r 18) has a mass of 18 g.

The large numbers in a balanced equation give the ratio of moles that react and form. To find a mass of product: change the known mass to moles, use the ratio from the equation, then change moles back to mass.

The limiting reactant is the one that is completely used up. It decides how much product forms. The other reactant is in excess and some of it is left over.

Rules to remember

number of moles = mass ÷ M_r (mass in g)
mass = number of moles × M_r
number of particles = number of moles × 6.02 × 10²³
The large numbers in a balanced equation are the mole ratio, e.g. 2Mg + O₂ → 2MgO means 2 mol Mg gives 2 mol MgO.
To find the limiting reactant, divide the moles of each reactant by its number in the equation; the smaller result is the limiting one.
To balance an equation from masses, change each mass to moles and find the simplest whole number ratio.
volume of gas at room temperature and pressure = number of moles × 24 dm³

Common mistake

Students multiply the M_r by the large number in the equation when working out moles. Use the M_r of one formula unit for moles, and use the large numbers only for the mole ratio.

Worked example

2Na + Cl₂ → 2NaCl. What mass of sodium chloride forms when 4.6 g of sodium reacts completely? (A_r: Na 23, Cl 35.5)

Moles of Na = 4.6 ÷ 23 = 0.20 mol.
Ratio Na : NaCl is 2 : 2, so 0.20 mol of NaCl forms.
M_r of NaCl = 23 + 35.5 = 58.5.
Mass of NaCl = 0.20 × 58.5 = 11.7 g.

Answer: 11.7 g of sodium chloride

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