Limiting and Excess Reactants: MDCAT Chemistry notes
Limiting and Excess Reactants for MDCAT: how to find the limiting reagent and calculate product formed and excess left, with worked examples.
Definitions
- Limiting reactant (reagent): the reactant that is consumed first. When it runs out the reaction stops, so it controls the amount of product. It is the reactant that gives the least amount of product.
- Excess reactant: the reactant present in more than the required amount; some of it is left over.
- Stoichiometric amount: the amount of a reactant called for by the balanced chemical equation.
An everyday picture: in the rusting of iron, oxygen and water from the air are effectively unlimited, so iron is the limiting reactant. Likewise, if you have 10 slices of bread and 3 eggs for sandwiches needing 2 slices and 1 egg each, the eggs limit you to 3 sandwiches.
Method
- Balance the equation.
- Convert each reactant's mass to moles ($n=m/M$).
- Calculate the amount of product expected from each reactant using the mole ratios. The reactant giving the smaller amount is limiting. (Equivalently, divide each reactant's moles by its coefficient; the smallest value is limiting.)
- Use the limiting reactant to find the product (theoretical yield).
- To find the excess left: moles of excess reactant used = moles of limiting × ratio; subtract from the amount taken.
Key formulas
$$n=\frac{m}{M}\qquad \text{excess left}=\text{amount taken}-\text{amount used}$$
$$\%\ \text{yield}=\frac{\text{actual}}{\text{theoretical}}\times100$$
Worked examples
Example 1. $2Mg+O_2\rightarrow2MgO$ with 50.0 g Mg ($M=24$) and 32 g $O_2$. Moles: Mg = 2.08, $O_2$ = 1.00. 1 mol $O_2$ needs 2 mol Mg = 48 g. Oxygen is limiting; 2 g Mg is left in excess, and 2 mol MgO = 80 g forms.
Example 2. $Ca+S\rightarrow CaS$ with 2.0 g Ca ($M=40$) and 4.0 g S ($M=32$). Moles: Ca = 0.050, S = 0.125. The ratio is 1:1, so Ca is limiting. CaS formed = 0.050 × 72 = 3.6 g; sulphur left = (0.125 − 0.050) × 32 = 2.4 g.
Example 3. $N_2+3H_2\rightarrow2NH_3$ with 56 g $N_2$ (2 mol) and 12 g $H_2$ (6 mol). The ratio 2:6 is exactly 1:3, so neither is in excess. Theoretical $NH_3$ = 4 mol = 68 g. If 51 g is obtained, % yield = $\frac{51}{68}\times100=75\%$.
Example 4. $2H_2+O_2\rightarrow2H_2O$ with 1 mol $H_2$ and 1 mol $O_2$: $H_2$ gives 1 mol water, $O_2$ could give 2 mol. $H_2$ is limiting; 0.5 mol $O_2$ is left over.
| Check | Limiting reactant | Excess reactant |
|---|---|---|
| Consumed | Completely, first | Partly |
| Product it could give | Least | More |
| Decides theoretical yield | Yes | No |
Common MDCAT traps
- The limiting reactant is not necessarily the one with the smaller mass; compare moles adjusted by coefficients.
- It gives the least product, not the most, and is used up first, not later.
- When the mole ratio exactly matches the equation, there is no excess reactant.
- Excess left should be reported for the excess reactant, e.g. "2 g Mg", not oxygen.
Quick revision
- Limiting reactant is consumed first and limits product.
- Calculate product from each reactant; the smaller answer wins.
- Stoichiometric amount comes from the balanced equation.
- In rusting, iron is limiting.