Stoichiometry

Limiting Reactant

The reactant which is totally consumed during the course of reaction and when it is consumed reaction stops.

The concept of limiting reactant is applicable to reaction other than monomolecular i.e., when more than one type reactant involved. For example

$2KCl{O}_3\to 2KCl+3O_2$. These is no limiting reactant.

To determine the limiting reagent amount of all reactants and mole ratio of reactants must be known. If the ratio of moles of reactant A with respect to reactant B is greater than the ratio of the moles of A to moles of B for a balanced chemical equation then B is the limiting reactant.

All other terms like left (unused) mass of other reactant, amount of formed product can be known stoichiometrically by knowing the amount of limiting reactant.

Method of Expressing Concentration of Solution

$\bullet$ Molarity (M): The molarity of a solution is the number of moles of solute present in one litre of the solution

$\text{M=}\frac{\text{w }\times \text{ 1000}}{{\text{m}}_{\text{1}}\times \text{ V}\text{in}\text{ml}}$

The molality (m): The molality is the number of moles of solute present in one Kg of solvent

$m=\frac{w_1\times 1000}{m_1\times w_2\left(ingrams\right)}$

Relation between molarity and molality

$m=\frac{1000\times M}{(1000\times d-M\times m_1)}$(Where d = density of solution)

Parts per million parts (ppm): For every dilute solution, i.e., when a very small quantity of a solute is present in large quantity of a solution, the concentration of the solute is expressed in terms of ppm. It is defined as the mass of the solute present in one million (10⁶) parts by mass of the solution. Thus for a solute A,

$\mathrm{pp}{\mathrm{m}}_{\mathrm{A}}=\frac{\mathrm{mass}\mathrm{of}\mathrm{A}}{\mathrm{mass}\mathrm{of}\mathrm{solution}}\times 10^6$

The pollution of the atmosphere is also reported in ppm but it is expressed in terms of volumes rather than masses, i.e. volume of the harmful gas (e.g. SO₂) in cm³ present in 10³ cm³ of the air.

(ii) Relationship between molarity (M) and molality (m): Molarity M means M moles of solute are present in 100 cc. of the solution. If density of the solution is d g/cc, mass of solution = 1000 f grams. Mass of solute = MM₂g (M₂ is mol mass of solute).

Hence mass of solvent = 1000d – MM₂g.

$\text{Molality}\left(\text{m}\right)\text{=}\frac{\text{M}}{\text{1000d-M}{\text{M}}_{\text{2}}}\times \text{1000}$

= $\frac{1000M}{1000d-M{M}_2}$

Thus m = $\frac{M}{d-M{M}_2/1000}$

or $m\left(d-\frac{M{M}_2}{1000}\right)=MorM\left(1+\frac{m{M}_2}{1000}\right)=md$

or $M=\frac{md}{1+m{M}_2/1000}$

(iii) Relationship between molality (m) and mole fraction (x₂): Molality (m) means m moles of the solute in 1000 g of the solvent = 1000 / M₁ moles (M₁ = mol of mass of the solvent). Hence

$Molefraction\left(x_2\right)=\frac{m}{m+1000/M_1}$

$=\frac{\mathrm{m}{\mathrm{M}}_1}{1000+\mathrm{m}{\mathrm{M}}_1}$

Thus $x_2=\frac{m{M}_1}{1000+m{M}_1}$

or $\frac{1}{x_2}=\frac{1000+m{M}_1}{m{M}_1}=\frac{1000}{m{M}_1}+1$

or $\frac{1000}{m{M}_1}=1-\frac{1}{x_2}=\frac{1-x_2}{x_2}=\frac{x_1}{x_2}$

or $m=\frac{1000x_2}{x_1{M}_1}$

(iv) Relationship between molarity (M) and mole fraction (x₂): Referring to calculations in (ii) above,

$Molefraction,x_2=\frac{M}{\left(1000d-M{M}_2\right)/M_1+M_2}$

$or{x}_2=\frac{M{M}_1}{\left(1000d-M{M}_2\right)+M{M}_1}$

= $\frac{M{M}_1}{M\left(M_1-M_2\right)+1000d}$

Thus x₂ = $\frac{M{M}_1}{M\left(M_1-M_2\right)+1000d}$

Or rearrangement, we get

Or $M=\frac{1000d{x}_2}{x_1{M}_1+x_2{M}_2}$

Note:

If molarity (M) is in moles / litre and density d is kg/litre and molality m is in moles / kg of the solvent, 1000 will be replaced by 1 in the above formulae.

Illustration:- Calculate the molarity of a solution containing 0.5 g of NaOH dissolved in

500 cm³.

Solution: Weight of NaOH dissolved = 0.5 g

Volume of the solution = 500 cm³

Calculation of molarity:

$0.5gofNaOH=\frac{0.5}{40}moleofNaOH$

$\left(\because Mol.massofNaOH=40\right)$

= 0.0125mole

Thus 500 cm³ of solution contain NaOH = 0.0125 mole

\ 1000 cm³ of the solution contain NaOH

= $\frac{0.0125}{500}\times 1000$ $\Rightarrow$ 0.025 mole

Hence molarity of the solution = 0.025 M