Stoichiometry
Introduced| English |
|---|
| mole ratio/məʊl ˈreɪʃɪəʊ/ |
| limiting reactant/ˈlɪmɪtɪŋ rɪˈæktənt/ |
The recipe math of chemistry
- A recipe says 2 eggs per cake -- chemistry has ratios too.
- A balanced equation gives the exact mole proportions.
- From how much you start with, you predict how much you will make.
- It is the arithmetic that runs every reaction.
Mole ratios from the equation
- The coefficients give the mole ratio 摩尔比 between substances.
- In $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$, that is 2 mol $\text{H}_2$ per 1 mol $\text{O}_2$.
- Use that ratio to convert moles of one into moles of another.
In $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$, how many moles of water form from 3 mol of $\text{O}_2$ (excess $\text{H}_2$)?
The ratio is 1 $\text{O}_2$ to 2 $\text{H}_2\text{O}$, so $3 \times 2 = 6\ \text{mol}$.
You must balance the equation before reading off the mole ratio.
The coefficients only give the correct ratio when balanced.
Grams to grams
- Convert grams to moles (divide by molar mass), apply the ratio, convert back.
- Grams, then moles, then moles, then grams -- the master path.
- Every stoichiometry problem walks this road.
Order the steps of a grams-to-grams stoichiometry problem.
Grams to moles, apply ratio, moles back to grams.
How many grams is 2 mol of water (molar mass 18)?
$m = nM = 2 \times 18 = 36\ \text{g}$.
The limiting reactant
- The limiting reactant 限量反应物 runs out first and caps the product.
- Compare the available moles against the ratio to find it.
- Whatever is left over is the excess reactant.
The stoichiometry roadmap
Convert a mass of one substance to a mass of another using the balanced equation.
The limiting reactant is the one that...
The reactant that runs out first caps how much product can form.
The reactant left over after the reaction stops is called the ____ reactant.
The non-limiting reactant is in excess.
How many moles of $\text{H}_2\text{O}$ form from $4\ \text{mol}$ of $\text{H}_2$ (with plenty of $\text{O}_2$)?
- The ratio $2\text{H}_2 \to 2\text{H}_2\text{O}$ is 1 : 1.
- So $4\ \text{mol}$ of $\text{H}_2$ gives $4\ \text{mol}$ of $\text{H}_2\text{O}$.
Always start from a balanced equation -- the coefficients are the mole ratio. Convert masses to moles before using the ratio, because you cannot compare grams directly. And the limiting reactant, not the bigger pile, sets how much product forms.
Stoichiometry uses the balanced equation's mole ratio to predict amounts. Convert grams to moles, apply the ratio, then convert back to grams. When reactants are both limited, the limiting reactant runs out first and caps the product; the rest is in excess.