Stoichiometry
1.0 The Mole
2.0 The Limiting Reagent
3.0 Gravimetric Analysis
4.0 Volumetric Analysis
5.0 Calculation of n-factor
6.0 Redox Reactions
7.0 Titration
7.1 Simple Titration
7.2 Double Titration
7.3 Method
7.4 Titration of the solution containing both $N{a_2}C{O_3}$ and $NaHC{O_3}$
7.5 Titration of the solution containing both $NaOH$ and $N{a_2}C{O_3}$
7.6 Back Titration
8.0 Iodimetric and Iodometric Titrations
9.0 Volume strength peroxide solution
10.0 Percentage Labeling of Oleum
11.0 Hardness of Water
1.1 Significance of Chemical Equations
7.2 Double Titration
7.3 Method
7.4 Titration of the solution containing both $N{a_2}C{O_3}$ and $NaHC{O_3}$
7.5 Titration of the solution containing both $NaOH$ and $N{a_2}C{O_3}$
7.6 Back Titration
Let us consider a balanced chemical equation
$PbS + 4{H_2}{O_2} \to PbS{O_4} + 4{H_2}O$
This equation will provide us various quantitative informations:
1. The molar ratio of reactants, i.e., $PbS$ and ${H_2}{O_2}$ in which they react together is 1:4.
2. The molar ratio of the two products i.e., $PbS{O_4}$ and ${H_2}O$ being formed in the reaction is also 1:4.
3. The initial moles of $PbS$ and ${H_2}{O_2}$ for the reaction to take place not necessarily be 1 and 4 respectively or also should not be in the molar ratio of 1:4.
4. One can start the reaction with $PbS$ and ${H_2}{O_2}$ in any molar ratio, but the ratio of $PbS$ and ${H_2}{O_2}$ which are reacting will always be in the ratio of 1:4.
5. One mole of $PbS{O_4}$ and 4 moles of ${H_2}O$ will be formed for each mole of $PbS$ being consumed.
The stoichiometric coefficients of a balanced chemical equation is the molar ratio and not the weight ratio.
One can use the balanced chemical equation for quantitative (gravimetric or volumetric) estimation of reactants and products. But if one does not have the balanced equation, it is very difficult to calculate the amounts of reactants consumed or products being formed.