This document provides an overview of chemical reactions and stoichiometry for a Grade 11 chemistry class. It discusses the balanced chemical equation for the formation of water, and how balancing the equation ensures the conservation of mass. Stoichiometry is introduced as the process of making calculations based on balanced chemical equations. Limiting and excess reactants are defined, with an example showing how to determine which reactant is limiting based on the mole ratios provided and actual amounts available.

1. Standard/ Class/ Grade XI Chemistry
Chapter 1 Basic Concepts
Gurudatta K Wagh, gkwagh@gmail.com
Chemical Reactions and Stoichiometry
Chemical reactions
Formation of water
hydrogen + oxygen → water H2(g) + O2(g) → H2O(g)
In this reaction one molecule (2 atoms) of hydrogen reacts
with one molecule (2 atoms) of oxygen to produce one
molecule of water containing two atoms of hydrogen and
one atom of oxygen
The mass of oxygen is not conserved

2. So we have to balance the equation,
2H2(g) + O2(g) → 2H2O(g)
Let us now consider the molar mass,
2 X 2.02 g + 1 X 32 g → 2 (2 X 1.01 + 16) g
4.04 g + 32 g → 2 (2.02 + 16) g
36.04 g → 36.04 g
Since the total mass of reactants = total mass of products,
the molar mass is conserved
Stoichiometry It is a process of making calculation based on
formulae and balanced chemical equations
Mass relationship The total mass of reactants must be
equal to the total mass of products

3. Limiting and excess reactants
In a chemical reaction the cost of the reactants plays an
important role. To save on the cost the cheaper reactant is
taken in excess and the costlier reactant is taken in lesser
quantity.
As the reaction begins the reactant that is taken in lesser
quantity (costlier reactant) gets consumed first and the
reaction stops. This reactant is called a limiting reactant.
On the other hand the reactant taken in excess (cheaper
reactant) does not get consumed fully and remains
unreacted with the product.

4. E.g. 2Au + 3Cl2 → 2AuCl3
2 X 196.97 g + 3 X 35.45 X 2 → 2 (196.97 + 3X 35.45)
393.94 g + 212.7 g → 606.64 g Hence 606.64 g →
606.64 g
molar ratio Au : Cl = 2 : 3 = 0.667
If 10 g each of Au and Cl2 is used
Mole of Au = 10 g/ 196.97 = 0.0508
Mole of Cl2 = 10 g/ 70.90 = 0.141
Ratio of actual moles of Au : Cl2 = 0.0508 : 0.141 =
0.36
Since we get a ratio less than 1, it means Cl2 is the
excess reactant and Au is the limiting reactant

5. Mole Au Mole Cl2
2 3
0.0508 ?
(0.0508 X 3) ÷ 2 = 0.0762
It means 0.0508 mole of Au required 0.0762 mole of Cl2
Mole of Cl2 that remains unreacted = 0.141 – 0.0762 = 0.0648
Mass of Cl2 that has reacted = 0.0762 X 70.90 = 5.402 g
Since 2 mole of Au produces 2 mole of AuCl3, hence
0.0508 mole of Au produces (0.0508 X 2) ÷ 2 = 0.0508 mole of
AuCl3
Mass of AuCl3 produced = 0.0508 X 303.32 (1 AuCl3 = 303.32 g)
= 15.409 g