Chemical Reactions and Stoichiometry

A balanced chemical equation is far more than a description of what reacts with what — its coefficients

are a precise numerical recipe, stating the exact ratio in which moles of reactants combine and moles of

products form. Stoichiometry is the branch of chemistry that uses this ratio to calculate quantities:

masses, moles, numbers of particles, or gas volumes involved in a chemical reaction.

Reading a balanced equation in moles. Consider the formation of ammonia:

This equation states that exactly $1\ \text{mol}$ of $\text{N}_2$ reacts with exactly $3\ \text{mol}$ of

$\text{H}_2$ to produce exactly $2\ \text{mol}$ of $\text{NH}_3$. These mole ratios — $1:3:2$ — hold true no

matter how much (or how little) ammonia is actually being made; they are fixed by the balanced equation, and

are the key to every stoichiometric calculation.

The general method. To calculate how much product forms from a given mass of reactant, or how much of one

reactant is needed to react completely with another:

1. Write and balance the chemical equation.
2. Convert the given mass of the starting substance into moles, using $n = \dfrac{m}{M}$ (Section 1.4).
3. Use the mole ratio from the balanced equation to find the number of moles of the substance you want to know about.
4. Convert that number of moles back into the required quantity — mass ($m = n \times M$), number of particles ($N = n \times N_A$), or gas volume at STP ($V = n \times 22.4\ \text{L}$).

Limiting reagent. In practice, reactants are rarely mixed in the exact mole ratio a balanced equation

demands. When two or more reactants are combined, whichever one runs out first — and therefore stops the

reaction, however much of the other reactant(s) remains unused — is called the limiting reagent. The

amount of product formed is always governed by the limiting reagent, never by the reactant present in excess.

To identify the limiting reagent:

1. Convert the given mass of each reactant into moles.
2. Divide each reactant's number of moles by its own coefficient in the balanced equation.
3. The reactant that gives the smaller value in step 2 is the limiting reagent — it is "used up" more completely relative to what the equation requires.
4. Use the limiting reagent's moles (via the mole ratio) to calculate the amount of product; the leftover quantity of the other reactant can also be calculated by subtracting what actually reacted from what was originally supplied.

For example, when hydrogen and oxygen react as $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g)$,

if the moles of $\text{H}_2$ available, divided by its coefficient $2$, gives a smaller number than the moles …