3.1 Mass-Mass Relationship

1.1 Significance of Chemical Equations

3.1 Mass-Mass Relationship
3.2 Mass-Volume Relationship
3.3 Volume-Volume Relationship

4.1 Titration
4.2 Molarity $(M)$
4.3 Molality $(m)$
4.4 Normality $(N)$
4.5 Equivalent Weight

5.1 Acid-Base Reaction

6.1 Precipitation/Double Decomposition Reactions

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

8.1 Iodimetric Titrations
8.2 Iodometric Titrations

11.1 Temporary Hardness
11.2 Permanent Hardness

It relates the mass of a species (reactant or product) with the mass of another species (reactants or product)

Let us consider a chemical reaction,

\[2NaHC{O_{3(s)}}\xrightarrow{\Delta }N{a_2}C{O_{3(s)}} + {\text{ }}{H_2}O{\text{ }} + {\text{ }}C{O_{2(g)}}\]

Suppose the mass of $NaHC{O_3}$ being heated is $‘a’$ gram and we want to calculate the weight of $N{a_2}C{O_3}$ being produced by heating of $‘a’$ gram of $NaHC{O_3}$.

The moles of $NaHC{O_3}$ $ = \frac{a}{{84}}$

According to the above balanced equation $2$ moles of $NaHC{O_3}$ upon heating gives $1$ mole of $N{a_2}C{O_3}$. So,

The no. of moles of $N{a_2}C{O_3}$ produced $ = \frac{1}{2} \times \frac{a}{{84}}$

Thus, wt. of $N{a_2}C{O_3}$ produced = moles of $N{a_2}C{O_3}$$ \times $ Molecular weight of $N{a_2}C{O_3}$

$$ = \frac{1}{2} \times \frac{a}{{84}} \times 106 = \frac{{53a}}{{84}}gms$$