Mole Ratios from Balanced Equations

At the particle level, $2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$ says that 2 hydrogen molecules collide and rearrange with 1 oxygen molecule to give 2 water molecules. Scale every coefficient by the same factor and the statement stays true: 2 dozen with 1 dozen gives 2 dozen, and 2 moles with 1 mole gives 2 moles, because a mole is just a (very large) count, $6.02\times 10^{23}$. That is why the coefficients can be read directly as a mole ratio. Grams are a different story: each substance has its own molar mass, so the same equation in grams reads $4.04\text{ g}$ of hydrogen with $32.00\text{ g}$ of oxygen giving $36.04\text{ g}$ of water, numbers you could never see in the coefficients.

This single multiplication is the pivot of every stoichiometry calculation you'll do this unit. Everything before it (converting a given mass to moles) and everything after it (converting the answer's moles to mass) are just applications of the mole map from earlier topics — the mole ratio itself is the only step that actually uses the chemical equation. The safest way to get the ratio the right way up is to write it as a conversion factor with full labels. The unit you have (mol of the given substance) must cancel, so it goes on the bottom, and the substance you want goes on top:

Any two substances in the equation can be paired, whether they are both reactants, both products, or one of each; the other substances simply don't enter the calculation. In ${\text{C}}_3{\text{H}}_8+5{\text{O}}_2\to 3{\text{CO}}_2+4{\text{H}}_2\text{O}$, oxygen and water are linked by $4:5$ even though propane is also present. The ratio also runs in either direction: going from ${\text{H}}_2\text{O}$ back to ${\text{O}}_2$ you multiply by $\frac{5}{4}$ instead of $\frac{4}{5}$. Because all the ratios come from one balanced equation, they are consistent with each other: $1.0\text{ mol}$ of propane uses $5.0\text{ mol}$ of oxygen and makes $3.0\text{ mol}$ of ${\text{CO}}_2$ and $4.0\text{ mol}$ of water, and you get the same numbers whichever substance you start from. Fractional answers are fine, since moles (unlike individual molecules) do not have to be whole numbers.

Build in a quick reasonableness check. If the wanted coefficient is larger than the given one, the answer must be more moles than you started with; if it is smaller, fewer. For ${\text{N}}_2+3{\text{H}}_2\to 2{\text{NH}}_3$, $3.0\text{ mol}$ of ${\text{H}}_2$ must give less than $3.0\text{ mol}$ of ${\text{NH}}_3$ (it gives $2.0\text{ mol}$). Because the number of particles is proportional to moles, the same ratio also works for molecules directly: 30 molecules of ${\text{H}}_2$ make 20 molecules of ${\text{NH}}_3$. Industry runs on these ratios. Ammonia plants that make fertilizer feed hydrogen and nitrogen in a $3:1$ ratio, and a car airbag holds sodium azide that decomposes by $2{\text{NaN}}_3\to 2\text{Na}+3{\text{N}}_2$, so every 2 mol of solid releases 3 mol of nitrogen gas to inflate the bag.