Limiting reactants (HT only)

• In many reactions, one reactant is deliberately added in excess to ensure the other is completely used up.

• The reactant that is used up first is called the limiting reactant because it limits the amount of product that can be formed.

• Once the limiting reactant is all used, the reaction stops – any other reactant left over is in excess.

Worked Example 1

Hydrochloric acid reacts with sodium hydroxide:

HCl + NaOH → NaCl + H₂O

Given:

36.5 g of HCl

80.0 g of NaOH

Step 1 – Calculate moles = mass / Mᵣ

Mᵣ of HCl = 36.5 ⇒ Moles of HCl = 36.5 ÷ 36.5 = 1.00

Mᵣ of NaOH = 40.0 ⇒ Moles of NaOH = 80.0 ÷ 40.0 = 2.00

Step 2 – Compare ratio

Equation ratio HCl : NaOH = 1 : 1

→ HCl is limiting (fewer moles than needed)

→ NaOH is in excess

Step 3 – Effect on product yield (determined by limiting reactant)

Because 1 mol of HCl produces 1 mol of NaCl, only 1.00 mol of NaCl can form.

Mass of NaCl = 1.00 × 58.5 = 58.5 g

Result: The maximum yield is 58.5 g of sodium chloride – limited by HCl.

Worked example 2 (challenging)

When the reactants don’t react in a 1:1 ratio, the number of moles can not be directly compared. Instead, the molar ratios from the balanced symbol equation are used to determine how much of each reactant is needed.

Sodium fluoride is formed when 4.60 g of sodium reacts with 5.70 g of fluorine.

Use calculations to show which reactant is in excess (4).

Calculate the maximum mass of sodium fluoride that can be formed (2).

1 . Calculate the moles of each reactant using moles =

2. Multiply the moles of the reactant with the smaller coefficient () by the larger coefficient ().

3. Identify whether you have enough (excess) or not enough (limiting).

4. Use the moles of the limiting reactant to determine the moles and mass of the product.

Recall

Application

Challenge (HT)