A balanced equation is a recipe, and its quantities are the big numbers written in front of the formulas: the coefficients. Take the neutralisation every plant runs, $2\,\mathrm{NaOH} + \mathrm{H_2SO_4} \rightarrow \mathrm{Na_2SO_4} + 2\,\mathrm{H_2O}$. It says two moles of caustic soda react with one mole of sulphuric acid. It says nothing about kilograms, and that is the trap this lesson is built around: the coefficients are a ratio of moles, never of masses.

$n_B = n_A\,\dfrac{b}{a}$, read aloud n-B equals n-A times b over a. State the convention once: A and B are the two substances being linked, and either one may sit on either side of the arrow. Every question says which is which. $n_A$ is the amount of A in moles and $n_B$ the amount of B in moles. $a$ is the coefficient written in front of A in the balanced equation and $b$ the coefficient in front of B. Both are bare whole numbers with no unit, and a formula with nothing in front of it carries a 1. Read forward, the relation gives $n_B$ from a known $n_A$. Read backward it is $n_A = n_B\,\dfrac{a}{b}$. The rule that never changes: the coefficient of what you want goes on top.

A balance reads kilograms, so the mole ratio usually arrives with a conversion on each side of it. $m_B = \dfrac{m_A}{M_A} \cdot \dfrac{b}{a} \cdot M_B$, read aloud m-B equals m-A over M-A, times b over a, times M-B. $m_A$ is the mass of A and $m_B$ the mass of B, both in the same unit, and you solve for whichever one the question leaves open. $M_A$ and $M_B$ are the two molar masses in grams per mole. Read it as three steps in a row: mass of A into moles of A, moles of A into moles of B, moles of B back into mass. Each molar mass stays with its own substance. Swapping them is the most common way this line goes wrong.

Work one to see the point. 49 kg of sulphuric acid at 98 g/mol is 500 mol. The equation asks for two moles of caustic soda per mole of acid, so 1000 mol. At 40 g/mol that is 40 kg. Apply the 2 straight to the kilograms instead and you would order 98 kg of caustic, nearly two and a half times too much.

Two cautions. The ratio is only true of a balanced equation. Read the coefficients off an unbalanced one and every number downstream is wrong by exactly the factor you failed to balance by, with nothing later in the calculation to flag it. And the answer assumes every other reactant is in excess, meaning there is more of it than the equation can use. When that is not true, something runs out early. That is a later lesson in this chapter.