Why the Ingredient That Runs Out First Decides How Much Product You Get

A cook with plenty of flour but only two eggs can bake only as many cakes as the eggs allow. Reactions work the same way, and chemists also need precise ways of stating how much solute a solution holds before they mix, dilute or titrate it.

This lesson covers limiting reagents and yield, concentration terms, and dilution and titration calculations.

The limiting reagent is the reactant used up first, which fixes the maximum amount of product — the theoretical yield — and the percentage yield compares the product actually obtained with that maximum.

Steps:

- Write the balanced equation
- Convert the given masses of reactants into moles
- Compare the mole ratio available with the ratio the equation requires
- The reactant that would run out first is the limiting reagent
- Use it to calculate the theoretical yield

Worked example. 28 g of nitrogen reacts with 9 g of hydrogen:

$${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2N{\mathrm{H}}_3$$

- Moles of nitrogen $=\frac{28}{28}=1$ mol; moles of hydrogen $=\frac{9}{2}=4.5$ mol
- 1 mol of nitrogen needs 3 mol of hydrogen, and 4.5 mol is available, so nitrogen is the limiting reagent
- Ammonia formed $=2$ mol $=2\times 17=34$ g, the theoretical yield
- Hydrogen left over $=4.5-3=1.5$ mol $=3$ g

Percentage yield:

$$\text{Percentage yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100$$

If only 27.2 g of ammonia is collected, the percentage yield is $\frac{27.2}{34}\times 100=80$.

An everyday example. Making sandwiches from 10 slices of bread and 3 slices of cheese gives only 3 sandwiches — the cheese runs out first, just like a limiting reagent.

The substance. The limiting reagent is not always the one with the smaller mass — here hydrogen has less mass, but nitrogen runs out first.

Molarity is moles of solute per litre of solution, molality is moles of solute per kilogram of solvent, normality is gram equivalents of solute per litre of solution, and mole fraction is the moles of one component divided by the total moles.

Definitions:

$$M=\frac{n_{\text{solute}}}{V_{\text{solution}}\text{ in L}}m=\frac{n_{\text{solute}}}{\text{mass of solvent in kg}}$$

$$N=\frac{\text{gram equivalents of solute}}{V_{\text{solution}}\text{ in L}}{x}_A=\frac{n_A}{n_A+n_B}$$

Worked example 1 — molarity. 4.0 g of NaOH (molar mass 40 g mol${}^{-1}$) is dissolved to make 250 mL of solution:

$$M=\frac{0.10\text{ mol}}{0.250\text{ L}}=0.40{\text{ mol L}}^{-1}$$

Worked example 2 — molality. 4.0 g of NaOH is dissolved in 500 g of water: $m=\frac{0.10\text{ mol}}{0.500\text{ kg}}=0.20$ mol kg${}^{-1}$.

Worked example 3 — mole fraction. 36 g of water (2 mol) is mixed with 46 g of ethanol (1 mol):

$$x_{\text{ethanol}}=\frac{1}{1+2}=0.33x_{\text{water}}=0.67$$

Worked example 4 — normality. Sulphuric acid gives 2 hydrogen ions per molecule (n-factor 2), so 0.5 M sulphuric acid is $0.5\times 2=1$ N.

Temperature. Molarity and normality change with temperature because volume changes, but molality and mole fraction do not.

An everyday example. Saline drips in Indian hospitals are prepared to an exact concentration so that the fluid entering a patient's blood matches the body's needs.

The substance. Molality uses the mass of the solvent, not of the whole solution — mixing the two up is a quick way to lose a mark.

**On dilution the moles of solute stay the same, so $M_1{V}_1=M_2{V}_2$, and in a titration the equivalents of acid and base that react are equal at the end point, so $N_1{V}_1=N_2{V}_2$.

Dilution:**

$$M_1{V}_1=M_2{V}_2$$

Worked example 1. 50 mL of 2.0 M hydrochloric acid is diluted to 500 mL:

$$M_2=\frac{2.0\times 50}{500}=0.20\text{ M}$$

Worked example 2. What volume of 10 M sulphuric acid is needed to make 1000 mL of 0.5 M acid?

$$V_1=\frac{0.5\times 1000}{10}=50\text{ mL}$$

Volumetric analysis (titration):

- A solution of known concentration is added from a burette to a measured volume of the unknown solution until the reaction is just complete
- An indicator such as phenolphthalein shows the end point by a colour change
- At the end point, equivalents of acid equal equivalents of base: $N_1{V}_1=N_2{V}_2$, or $M_1{V}_1=M_2{V}_2$ for a 1 : 1 reaction

Worked example 3. 25.0 mL of sodium hydroxide solution is neutralised by 20.0 mL of 0.10 M hydrochloric acid:

$$M_{\text{NaOH}}=\frac{0.10\times 20.0}{25.0}=0.080\text{ M}$$

Worked example 4. 20 mL of 0.1 M sulphuric acid is 0.2 N, so the volume of 0.1 M NaOH it neutralises is

$$V_{\text{NaOH}}=\frac{0.2\times 20}{0.1}=40\text{ mL}$$

An everyday example. Food testing laboratories titrate vinegar against sodium hydroxide to check that its acetic acid content is correct.

The substance. Always add concentrated acid to water, never water to acid — dilution releases heat, and adding water to acid can make it splash dangerously.