.STOICHIOMETRY AND VOLUMETRIC

The word stoichiometry is derived from two Greek words - stoicheion (meaning element) and metron (meaning measure).

Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction.

How many moles of methane are required to produce 22g CO2 (g) after combustion?

Solution

According to the chemical equation.

CH4 (g) +2O2 (g) → CO2 (g) + 2H2O (g)

44g CO2 (g) is obtained from 16 g CH4 (g).

1 mol CO2 (g) is obtained from 1 mol of CH4 (g)]

 

= 0.5 mol CO2 (g)

Hence, 0.5 mol CO2 (g) would be obtained from 0.5 mol CH4 (g) or 0.5 mol of CH4 (g) would be required to produce 22 g CO2 (g)

Ex.:- At 300 K and I atmospheric pressure, 10mL of a hydrocarbon required 55mL of O2 for complete combustion and 40 mL, of CO2 is formed The formula of the hydrocarbon is        (2019 Main, 10 April)

(a) C4H7Cl                                             (b) C4H6

(c) C4H10                                               (d) C4H8

Solution: In eudiometry,

  

 

 

 

Given, (i) VCO2 = 10x = 40mL Þ x = 4

 

 

          

 

 

So, the hydrocarbon (Cx Hy) is C4H6         Answer (b)

LIMITING REAGENT

Many a time, the reactions are carried out when the reactants are not present in the amounts as required by a balanced chemical reaction. In such situations, one reactant is in excess over the other. The reactant which is present in the lesser amount gets consumed after sometime and after that no further reaction takes place whatever be the amount of the other reactant present. Hence, the reactant which gets consumed, limits the amount of product formed and is, therefore, called the limiting reagent.

IMPORTANT: Product formed is calculated by limiting reagent

Limiting reagent

2A + 3B → 4C

If we have 3 mole of A of 4 mole of B, find out mole of C farmed

2A + 3B → 4C

   3      4

3/2x  3 = 4.5 > 4 so B is limiting reagent

3 mole of B form 4 mole C

4 mole of B will form

       Answer: 5.33 moles

Ex.:- If 0.50 mole of BaCl2 is mixed with 0.20 mole of Na3PO4, the maximum number of moles of Ba3 (PO4)2 that can be formed is                                                                                      (1981, 1M)

(a) 0.70                                                 (b) 0.50

(b) 0.20                                                 (d) 0.10

Solution: The balanced chemical reaction is

In this reaction, 3 moles of BaCl2 combines with 2 moles of Na3PO4. Hence, 0.5 mole of BaCl2 require

 

Since, available Na3PO4 (0.2 mole) is less than required mole (0.33), it is the limiting reactant and would determine the amount of product Ba3(PO4)2

2 moles of Na3PO4 gives 1 mole Ba3 (PO4)2

       

 Answer (d)