Reacting Quantities

This lesson covers: 

1. What the limiting reactant is
2. How to identify the limiting reactant
3. Calculating amounts of products based on the limiting reactant

The limiting reactant limits the amount of product

When two or more reactants are mixed together, one reactant will always be completely used up before the others.

• The reactant that runs out first is called the limiting reactant.
• Any reactants present in excess are called excess reactants.
• The amount of limiting reactant present sets an upper limit on how much product can be formed.

For example, in the reaction:

A + B ➔ C

If 1 mole of A reacts with 2 moles of B, then A will be the limiting reactant as B is present in excess.

So if the reaction proceeds to completion, 1 mole of A will produce 1 mole of C, leaving 1 mole of B unreacted.

How to identify the limiting reactant

There are two steps to determining which reactant is limiting:

1. Calculate the number of moles of each reactant present.
2. Compare the number of moles of each reactant to the stoichiometric ratios in the balanced chemical equation:

• The reactant for which the available number of moles is lower relative to the stoichiometric ratio is the limiting reactant.
• The other reactants, for which there are excess moles beyond the stoichiometric ratio, are in excess.

For example, consider the reaction between hydrogen and oxygen to form water:

2H_2(g) + O_2(g) ➔ 2H₂O_(l)

According to the balanced equation, the stoichiometric ratio is 2 moles of H₂ per 1 mole of O₂.

If equal 1 mole quantities of H₂ and O₂ react, there would only be 1 mole of H₂ per 1 mole O₂, instead of the required 2:1 ratio. Since the amount of H₂ is lower than the stoichiometric ratio relative to O₂, hydrogen would be the limiting reactant.

Worked example 1 - Determining the limiting reactant

23.0 g of C₂H₅OH undergoes complete combustion with 50.0 g of O₂. The balanced equation is: 

C₂H₅OH_(l) + 3O_2(g) ➔ 2CO_2(g) + 3H₂O_(g)

Determine the limiting reactant.

Step 1: Calculate number of moles of each reactant

Moles of C2​H5​OH =Mr ​m​=46.023.0​=0.50 mol

Moles of O2​ =Mr ​m​=32.050.0​=1.56 mol

Step 2: Compare to stoichiometric ratio (1:3)

For 0.50 mol of C₂H₅OH, 0.50 x 3 = 1.50 mol of O₂ is required

Step 3: Compare number of O₂ moles required vs. present

Moles of O₂ required = 1.50 mol

Moles of O₂ present = 1.56 mol

Step 4: Determine the limiting reactant

1.56 > 1.50 so O₂ is in excess and C₂H₅OH is the limiting reactant.

Basing calculations on limiting reagent

The maximum amounts of products formed depends on the limiting reactant:

• Amounts of products can only be calculated based on moles of limiting reactant.
• Excess reactants does not affect product amounts

For example, in the reaction:

C₂H₅OH_(l) + 3O_2(g) ➔ 2CO_2(g) + 3H₂O_(g)

0.50 mol C₂H₅OH can produce:

• 0.50 × 2 = 1.0 mol of CO₂
• 0.50 × 3 = 1.50 mol H₂O

So the maximum amounts of carbon dioxide and water produced are 1.0 mol and 1.50 mol respectively, regardless of O₂ in excess.

Performing calculations based on limiting reagent is essential to determine actual yields.

Worked example 2 - Using the limiting reactant to calculate mass of product

Calculate the mass (in g) of sodium chloride (NaCl) that can be produced when 200 cm³ of 1.0 mol dm^-3 hydrochloric acid (HCl) reacts with 150 cm³ of 1.2 mol dm^-3 sodium hydroxide (NaOH). The balanced equation is:

HCl_(aq) + NaOH_(aq) ➔ NaCl_(aq) + H₂O_(l)

Step 1: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

200 cm³ = 0.200 dm³

150 cm³ = 0.150 dm³

Step 2: Calculate number of moles of each reactant

Moles of HCl = c × V = 1.0 x 0.200 = 0.200 mol

Moles of NaOH = c x V = 1.2 x 0.150 = 0.180 mol

Step 3: Compare to stoichiometric ratio (1:1)

For 0.200 mol of HCl, 0.200 mol of NaOH is required

Step 4: Compare number of NaOH moles required vs. present

Moles of NaOH required = 0.200 mol

Moles of NaOH present = 0.180 mol

Step 5: Determine the limiting reactant

0.180 < 0.200 so NaOH is the limiting reactant and HCl is in excess.

Step 6: Calculate number of moles of NaCl formed

NaCl : NaOH mole ratio = 1:1

Moles of NaCl = 0.180 mol

Step 7: Calculate mass of NaCl formed

m = n x M_r = 0.180 x 58.5 = 10.5 g

Therefore, 10.5 g of sodium chloride is produced with sodium hydroxide acting as the limiting reactant.

Worked example 3 - Using the limiting reactant to calculate mass of product

Calculate the mass (in g) of titanium metal (Ti) that can be produced when 500 g of titanium tetrachloride (TiCl₄) reacts with 100 g of magnesium metal (Mg). The balanced equation is:

TiCl_4(l) + 2Mg_(s) ➔ Ti_(s) + 2MgCl_2(s)

Step 1: Calculate number of moles of each reactant

Moles of TiCl4​ =Mr ​m​=189.9500​=2.63 mol

Moles of Mg =Ar ​m​=24.3100​=4.12 mol

Step 2: Compare to stoichiometric ratio (1:2)

For 2.63 mol of TiCl₄, 2.63 x 2 = 5.26 mol of Mg is required

Step 3: Compare number of Mg moles required vs. present

Moles of Mg required = 5.26 mol

Moles of Mg present = 4.12 mol

Step 4: Determine the limiting reactant

4.12 < 5.26 so Mg is the limiting reactant and TiCl₄ is in excess.

Step 5: Calculate number of moles of Ti formed

Ti : Mg mole ratio = 1:2

Moles of Ti =24.12​=2.06 mol

Step 6: Calculate mass of Ti formed

m = n x A_r = 2.06 x 47.9 = 98.6 g

Therefore, 98.6 g of titanium metal can be produced, with magnesium acting as the limiting reactant.