This document provides examples of how to calculate the limiting reactant and maximum product formed in chemical reactions. It demonstrates calculating moles of each reactant, comparing reactant ratios to determine the limiting reactant, and using mole ratios to calculate maximum product. An alternative method is proposed to systematically solve these types of problems.

1. Limiting reactants
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2. Problem solving
Aim: Be able to calculate:
A. Which reactant (if any) is limiting?
B. What mass or volume of product can be formed?

3. A. Which reactant is limiting?
750g of ammonia and 750 g of
oxygen are combined.
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
1 Calculate moles of each reactant:
moles = mass/mr
Moles NH3 = 750 / 17.03 = 44 moles Moles O2 = 750 / 32 = 23.4
2 Check the required ratio compared to the actual ratio.
Require 5mole O2 / 4mole NH3 = 1.25moles O2 / 1mole NH3
Available 23.4mole O2 /44mole NH3 = 0.523moles O2 / 1 mole NH3
Oxygen is the limiting reagent!

4. B. What mass of NO can we form?
750g of ammonia and oxygen are
combined.
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
1. Work out moles of limiting reactant
Previous slide moles O2 = 23.4
2 Use the product / reactant mole ratio to work out moles of product
23.4 x (4/5) = 18.72 moles of NO
3 Convert moles to mass
Mass = moles x mr = 18.72 x 30.1 = 562g of NO

5. 125g of Cl2 and 125g of C were added to an excess of
TiO2 and the following reaction occurred:
TiO2(s) + 2Cl2(g) + C(s) → TiCl4(l) + CO2(g)
A. Which reactant is limiting?
1 Calculate moles of each reactant:
moles = mass/mr
Moles Cl2 = 125 / 71 = 1.76 moles Moles C = 125 / 12 = 10.4 moles
2 Check the required ratio compared to the actual ratio.
Require 2moles Cl2 / 1mole C
Available 1.76moles Cl2 /10.4 moles C = 0.17 Cl2 / 1 mole C
Cl2 is the limiting reagent

6. 125g of Cl2 and 125g of C were added to an excess of
TiO2 and the following reaction occurred:
TiO2(s) + 2Cl2(g) + C(s) → TiCl4(l) + CO2(g)
B What mass of TiCl4 can be produced?
1. Work out moles of limiting reactant
Previous slide Cl2 = 1.76 moles
2 Use the product / reactant mole ratio to work out moles of product
1.76 x (1/2) = 0.88 moles of TiCl4
3 Convert moles to mass
Mass = moles x mr = 0.88 x 189.7 = 167g of TiCl4

7. Practice
1. 356g of CO and 65g of H2 react according to the equation:
CO(g) + H2(g) → CH3OH(l)
Which reactant is limiting and what is the maximum mass of methanol that can
be formed?
2. The thermite reaction is used to produce molten Iron:
Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s)
50g of Al and 50g Fe2O3 is used. Which reactant is limiting? What is the
maximum mass of iron that can be produced?
3. A 2l bottle contains 800cm3
of propane and the remainder is oxygen.
C3H8 + 5O2 → 3CO2 + 4H2O
Which reactant is limiting? What is the maximum volume of CO2 that can be
produced? Why might a greater volume of CO2 actually be produced?!?

8. Answers
1. CO is limiting and 407g of CH3OH can be produced
2. Fe2O3 is limiting and 35g of Fe can be produced
3. O2 is limiting and 0.72dm3
(= 720cm3
) of CO2 could be formed.
Exothermic reaction causes expansion of gases.

9. Extension
Design your own question on limiting
reactants.
Develop a different method for solving these
types of problems that will work in every
case.

10. Extension
Design your own question on limiting
reactants.
Develop a different method for solving these
types of problems that will work in every
case.