Mole Ratio from a Balanced Equation

This is the hinge every stoichiometry problem turns on, and it is one line long. The coefficients in a balanced equation are a ratio of moles — never of masses, never of volumes, never of anything you can weigh directly. So the trip from one substance to another has to pass through moles, and the mole ratio b/a is the bridge.

Worked: the Haber process, N₂ + 3H₂ → 2NH₃. Feed 6.00 mol of hydrogen and ask how much ammonia it can make. The coefficient on hydrogen is 3, on ammonia 2, so n(NH₃) = 6.00 × 2/3 = 4.00 mol. Not 6.00, not 9.00 — the equation says three hydrogens buy two ammonias, and no amount of arithmetic elsewhere changes that exchange rate.

The failure mode is reading coefficients off an unbalanced equation, and nothing downstream will catch it. Write N₂ + H₂ → NH₃ and the ratio looks like 1:1; every mass, every volume, every yield you compute afterwards is then wrong by a clean factor, and every one of them will look perfectly reasonable. Balance first, count atoms on both sides, and only then read the numbers.

Two smaller traps. First, the ratio is not symmetric — going from A to B multiplies by b/a, and going back multiplies by a/b, so the direction of the arrow matters as much as the numbers. Second, a balanced equation can legitimately be written with fractional coefficients, as in H₂ + ½O₂ → H₂O. That is fine and the ratio still works; it just means the ΔH quoted alongside belongs to that version of the equation and not to the doubled one.

Everything else in a mass-to-mass calculation is unit conversion. Molar mass turns grams into moles on the way in, this ratio does the chemistry in the middle, and molar mass turns moles back into grams on the way out. Students who find stoichiometry hard almost always find the middle step easy and the surrounding conversions hard; the ratio itself is the part with the chemistry in it.