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

${\text{Moles of C}}_2{\text{H}}_5\text{OH }=\frac{\text{m}}{{\text{M}}_{\text{r }}}=\frac{23.0}{46.0}=0.50\text{ mol}$

${\text{Moles of O}}_2=\frac{\text{m}}{{\text{M}}_{\text{r }}}=\frac{50.0}{32.0}=1.56\text{ mol}$

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

For 0.50 mol of C₂H₅OH, 0.50 x 3 = 1.50 mol of O_2_ 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 (TiCl_4_) 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

${\text{Moles of TiCl}}_4=\frac{\text{m}}{{\text{M}}_{\text{r }}}=\frac{500}{189.9}=2.63\text{ mol}$

$\text{Moles of Mg }=\frac{\text{m}}{{\text{A}}_{\text{r }}}=\frac{100}{24.3}=4.12\text{ mol}$

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

$\text{Moles of Ti }=\frac{4.12}{2}=2.06\text{ mol}$

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.

Worked example 4 - Using the limiting reactant to calculate mass of product

1.20 dm³ of ammonia gas (NH₃) reacts with 1.00 dm³ of hydrogen chloride gas (HCl) to form ammonium chloride (NH_4_Cl).

Calculate the mass of ammonium chloride produced, assuming complete reaction (all gas volumes measured at r.t.p). The balanced equation is:

NH_3(g)_ + HCl_(g)_ ➔ NH_4_Cl_(s)_

Step 1: Calculate number of moles of each reactant

At r.t.p, 1 mole of gas occupies 24.0 dm³

${\text{Moles of NH}}_3=\frac{\text{V}}{{\text{V}}_{\text{m }}}=\frac{1.20}{24.0}=0.0500\text{ mol}$

$\text{Moles of HCl }=\frac{\text{V}}{{\text{V}}_{\text{m }}}=\frac{1.00}{24.0}=0.0417\text{ mol}$

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

For 0.0500 mol of NH₃, 0.0500 mol of HCl is required

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

Moles of HCl required = 0.0500 mol

Moles of HCl present = 0.0417 mol

Step 4: Determine the limiting reactant

0.0417 < 0.0500 so HCl is the limiting reactant and NH₃ is in excess.

Step 5: Calculate number of moles of NH₄Cl formed

HCl : NH_4_Cl mole ratio = 1:1

Moles of NH_4_Cl = 0.0417 mol

Step 6: Calculate mass of NH_4_Cl formed

m = n x M_r = 0.0417 x 53.5 = 2.23 g

Therefore, 2.23 g of ammonium chloride can be produced, with hydrochloric acid acting as the limiting reactant.