Limiting Reagents/Reactants

• Limiting Reactant- the reactant that is completely consumed in the chemical reaction

• Excess Reactant- the reactant that is present in more than required quantities in the chemical reaction

• When the limiting reactant is used up, the reaction will stop. No more product can be made, regardless of how much of the excess reactant may be present.

• Therefore, the limiting reactant determines how much product is produced

Real Life Examples• Washing dishes – you can have too much, or

too little soap or grease

• Sugar in blood levels in your body. Too much sugar you are tired, too little, have no energy

• Salt & Potassium in body. Muscle cramps & thirst

Solving Limiting Factor Problems

When you are given amounts of two or more reactants to solve a stoichiometric problem, you must identify the limiting reactant

How to identify the limiting reactant:• Problem type: What mass of product C could

be obtained when mass A is reacted with mass B?

• Chemical reaction format: 2A+ B 3C + D

Problem Solving Strategy

Mol A

Mass C

Mol C

Mass AMass B

Mol A Mol B

2A+ B 3C + D

Note: 2 masses given

Use mole ratio to determine the limiting factor(e.g. A is limiting)

The limiting factor always controls the amount of product being produced

Mole ratio

Example 1: A reaction contains 134.9g of Aluminium and 96.0g of oxygen. Determine the

excess & limiting reagent and the amount of product formed.

• Al + O2 Al2O3

• 4Al + 3O2 2Al2O3

Step 1: Write out the balanced chemical equation

• Step 2: Fill in chart with information you know

Balanced

Equation4Al 3O2

2Al2O3

Mole Ratio

4 3 2

Mass (m) 134.9g 96.0gMolar Mass (M)

26.98g/mol 32.0g/mol

Moles (n)

Step 3: Convert given mass into moles for both (n=m/M)

Moles of O2 = 96.0g

32.0g/mol= 4.99 mol of Al

Moles of Al = 134.9g

27.0g/mol

= 3.00 mol of O2

• Fill in chart with information you know

Balanced

Equation4Al 3O2

2Al2O3

Mole Ratio

4 3 2

Mass (m) 134.9g 96.0g ?g

Molar Mass (M)

26.98g/mol 32.0g/mol

Moles (n) 4.99mol 3.00mol

Step 4: Take the moles of each reactant and using the mole ratio, determine how much product would be made

The O2 limiting reactant because it produces the least amount of moles of product (Al2O3)

Mols of Al2O3 produced from O2

3.00 mol O22 mol Al2O3

3 mol O2 X =

2.00 mol Al2O3

Mols of Al2O3 produced from Al

4.99 mol Al 2 mol Al2O3

4 mol Al X = 2.50 mol Al2O3

• Fill in chart with information you know

Balanced

Equation4Al 3O2

2Al2O3

Mole Ratio

4 3 2

Mass (m) 134.9g 96.0g ?g

Molar Mass (M)

26.98g/mol 32.0g/mol 102g/mol

Moles (n) 4.99 mol 3.00mol 2.0mol

Step 5: Convert moles of required substance to the mass (m=n x M)

mass of Al2O3 = 2.00 mol of Al2O3 X 102g/mol

= 204g of Al2O3

Example 2: Pentane (C5H12) is a major component of gasoline. What mass of water would be produced when

28.5 g of pentane reacts with 3.00 g of oxygen gas?

1C5H12 + 8O2 6H2O + 5CO2

Reactants Products

m= 28.5g m= 3.00g m= ?g

• Step 2: Fill in chart with information you know

Balanced

Equation 1C5H12 8O2 6H2O 5CO2

Mole Ratio

1 8 6 5

Mass (m) 28.5g 3.00g

Molar Mass (M)

72.17g/mol 32g/mol

Moles (n)

Step 3: Convert given mass into moles for both (n=m/M)

Moles of O2 = 3.00g 32g/mol

= 0.3958 mols of C5H12

Moles of C5H12 =

= 0.09375 mol of O2

28.5g72g/mol

• Fill in chart with information you know

Balanced

Equation 1C5H12 8O2 6H2O 5CO2

Mole Ratio

1 8 6 5

Mass (m) 28.5g 3.00g

Molar Mass (M)

72.17g/mol 32g/mol

Moles (n) 0.3958 0.09375

Step 4: Take the moles of each reactant and using the mole ratio, determine how much product would be made

The O2 limiting reactant because it produces the least amount of moles of product (H2O)

Mols of H2O produced from O2

0.09375 mol O26 mol H2O

8 mol O2 X =

0.703 mol H2O

Mols of H2O produced from C5H12

0.3958 mol C5H126 mol H2O

1 mol C5H12 X = 2.37 mol H2O

• Fill in chart with information you know

Balanced

Equation 1C5H12 8O2 6H2O 5CO2

Mole Ratio

1 8 6 5

Mass (m) 28.5g 3.00g

Molar Mass (M)

72.17g/mol 32g/mol 18.02g/mol

Moles (n) 0.09375 0.3958 0.703 mols

Step 5: Convert moles of required substance to the mass (m=n x M)

mass of H2O = 0.0703 mols H2O X 18.02g/mol

= 1.27g of H2O

Example 3: Calculate the mass of aluminium chloride that can be

produced from 20.0 g of aluminium and 30.0 g of chlorine gas.

Step 1: Write the balanced equation for the reaction.

2 Al (s) + 3 Cl2(g) 2 AlCl3(s)

Step 2: Calculate the number of moles of each reactant. (Using n=m/M) 2 Al (s) + 3 Cl2(g) 2 AlCl3(s)

m 20.0 g 30.0gM 27.0g/mol 2(35.5) =71.0g/moln = 0.741 mol Al =0.423 mol Cl2

20.0g27.0g/mol

30.0g71.0 g/mol

Step 3: Take the moles of each reactant and using the mole ratio, determine how much product would be made

The Cl2 limiting reactant because it produces the least amount of moles of product (AlCl3)

Mols of AlCl3 produced from Al

0.741 mol Al 2 mol AlCl3

2 mol Al X =

0.741 mol AlCl3

Mols of AlCl3 produced from Cl2

0.423 mol Cl22 mol AlCl3

3 mol Cl2

X = 0.282 mol AlCl3

Step 4. Calculate the number of

moles using the limited reagent.

Moles of AlCl3 = nM

= 0.423mol(133.33g/mol)

= 56.4 g of AlCl3