Standard/ Class/ Grade XI ChemistryChapter 1 Basic ConceptsGurudatta 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

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) g4.04 g + 32 g → 2 (2.02 + 16) g36.04 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

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.

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 = 0.667If 10 g each of Au and Cl2 is usedMole of Au = 10 g/ 196.97 = 0.0508Mole of Cl2 = 10 g/ 70.90 = 0.141Ratio 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

Mole Au Mole Cl2

2 30.0508 ?(0.0508 X 3) ÷ 2 = 0.0762It means 0.0508 mole of Au required 0.0762 mole of Cl2

Mole of Cl2 that remains unreacted = 0.141 – 0.0762 = 0.0648Mass of Cl2 that has reacted = 0.0762 X 70.90 = 5.402 g

Since 2 mole of Au produces 2 mole of AuCl3, hence0.0508 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