شرح درس
The Mole and Avogadro's number
CHEMISTRY - Secondary 1
In This part we will study
The limiting reactant in the reaction
To clarify and simplify the concept of the limiting reactant, the following application will be studied :
When you buy an egg carton, you will find that this one egg carton contains 30 eggs , this can be expressed by the following equation :
* Can you determine mathematically the number of egg cartons which can be prepared from :
• 150 eggs.
• 4 empty cartons.
Which of them its number exceeds the required number for the filling process , eggs or empty cartons ?
* This can be concluded through the previous equation, as follows:
Number of eggs + Number of empty cartons → Number of egg cartons

*So, it is obvious that to use all the eggs in the filling, we need 5 empty cartons while the available empty cartons are only 4, therefore not all the eggs will be used but all the empty cartons will be used to produce egg cartons in a number that is lower than the number of the egg cartons which would be produced if we used all eggs, this is why we call the empty cartons "the limiting reactant" of the process of filling.
Number of used eggs and the number of remaining eggs can be calculated , as follows :
The limiting reactant : The reactant which is completely consumed during the chemical reaction and reacts with other reactants to produce the lowest number of moles of the product.
Example (1) :
Potassium hydroxide solution reacts with sulphuric acid according to following equation :
2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)
What is the limiting reactant, when 4 mol of sulphuric acid are present with 3 mol of potassium hydroxide in the reaction medium ?

Solution :
Example (2) : Choose the correct answer :
7 g of iron react with 4 g of sulphur and produces 11 g of iron (II) sulphide, What will remain in the reaction container when 7 g of iron are added to 7 g of sulphur ? ...........
14 g of iron (II) sulphide only. 11 g of iron (II) sulphide, 3 g of iron.
11 g of iron (II) sulphide only. 11 g of iron (II) sulphide, 3 g of sulphur.
Idea of Solution
- Each 7 g of Fe react completely with 4 g of S to form 11 g of FeS
- 11 g of FeS will be formed with remaining of (7 - 4 = 3 g) of unreacted sulphur.
So ; Choice is ()
Example (3) :
Hydrogen gas reacts with oxygen gas to form water according to the equation :
2H2(g) + O2(g) → 2H2O(v)
Calculate the mass of water produced from the reaction of 16 g of O2 with 16 g of H2 [H = 1,0=16]
Solution:
Mass of produced water on consuming:

- The lower mass of water is produced as a result of consuming all the mass of O2
- O2 is the limiting reactant.
- Mass of produced water= 18 g

Best wishes to you
Mr.Ahmed Elbasha

كيمياء الصف الاول الثانوي لغات - الفصل الدراسي الاول
شرح كيمياء لغات اولي ثانوي - مستر احمد الباشا
كيمياء لغات - الصف الاول الثانوي - كيمياء اولي ثانوي الترم الاول - كيمستري اولي ثانوي الترم الاول