SOLVED NUMERICALS ON MOLE CONCEPT AND POAC :- BOOST UP YOUR CONCEPT ABOUT "APPLICATION OF POAC AND MOLE CONCEPT"

Q.1 what amount of CaO will be produced by 1 gm. of Ca?

This numerical we can solve by two methods

 1.     Mole concept  2. POAC method

Advantage of POAC over Mole concept is that we need not to write down the stoichiometry of reaction . means in POAC method NEED NOT to  BALANCE THE CHEMICAL REACTIONS

By mole concept

First wite down the BALANCE CHEMICAL REACTION

2Ca + O₂ → 2CaO

It means when 2mole of ca ( 2 *40=80 gm) react with 1 mole of oxygen it produce 2 

mole of CaO (2*56= 112 gm).

80 gm Ca produce = 112 gm Cao

So 1 gm Ca will produce = 112/ 80 = 1.4 gm CaO (unitary method)

By POAC (ADAVANTAGE OVER mole concept NO NEED TO BALANCE EQUATION)

Ca + O₂ → CaO

 Apply POAC on Ca

Total no. of moles of Ca atoms in reactant side = Total no. of moles of Ca atoms in product side

1* no. of moles of Ca = 1* no. of moles of CaO

 No. of moles of Ca /  no. of moles of CaO = 1

So mole of CaO produced in reaction would be equal to mole of Ca  used in chemical reaction

because we are using 1 gm Ca means ( no. of mole =mass/ atomic wt ; here atomic wt

 of Ca = 40 ; so  no. moles of Ca =1/40 =0.025 moles)  0.025  moles of Ca ,the total moles of Cao would be same means 0.025 moles of Cao means (0.025 * 56 = 

1.4) 1.4 gm Cao . Here m.wt of CaO =56 gm

POAC method is a good method to solve these type of numerical because you can  apply POAC concept directly without balancing of equation. So u can save your time  in competitive exams. Even in starting students may feel to apply POAC something difficult but actually it is better than mole concept and time saving method. So students  have to be practice to solve this type of numerical by POAC method. Here some numericals are solved by POAC method to clarify your concept about POAC .observe these numericals to improve your understanding about POAC.

Q.2 A sample of  KClO₃ on decomposition yielded 448 ml.of oxygen gas at N.T.P. calculate

i. weight of oxygen produced

ii. weight of KClO₃ originally taken, and

iii. wt. of KCl produced

solution:-

i. Calculation of  Wt. of  O₂

  VOL. of O₂ = 448 ml = 0.448 litre

  Mole of  O₂ = 0.448/22.4 = 0.02 moles means

  Mass = no. of mole *  m.wt = 0.02 * 32 = 0.064 gm

ii.Calculation of  Wt of KClO₃

KClO₃  → KCl + O₂
Apply POAC on O atom

3* no. of moles of KClO₃ = 2* no. of moles of O₂

No. of moles of KClO₃ / No. of moles of O₂ = 2/3

No. of moles of KClO₃ = 2/3 * no. of moles O₂

No. of moles of KClO₃ = 2/3 * 0.02 = 0.0133 moles

Mass of  KClO₃ = 0.0133 * 122.5 = 1.633 gm  

Here 122.5 is the m.wt of  KClO₃

iii.Calculation of wt. of  KCl produced 

Apply POAC on K atoms

1* no. of mole of KClO₃ = 1* no. of mole of KCl

So no. of mole of  KCl = NO. of mole of KClO₃ = 0.0133 moles

Mass of KCl = 0.0133 * 74.5 = 0.99 gm

Here 74.5 is the m.wt. of  KCl 

Q.3 A mixture of  KBr and NaBr weighing 0.560 gm was treated with aq. Ag+ ion and all bromide ion was recovered as 0.970 gm of pure AgBr. What was the friction by weight of KBr in the sample?

Solution :- ( KBr   +   NaBr)   + Ag⁺ → AgBr

              (x gm      0.560 –x gm)                                0.970 gm

Because no. of moles of Br atoms are conserved in reaction so

Apply POAC on Br atom

Moles of Br in KBr + moles of Br in NaBr = moles of Br in AgBr 

1* no.of mole of KBr + 1*no. of mole of NaBr =   1* no. of mole of AgBr

Mass of KBr/ m.wt of KBr + Mass of NaBr / M.wt of NaBr  = Mass of AgBr/ M.wt  of AgBr

x/ 119 + 0.560- x / 103 = 0.97/188

x = 0.1332 gm

So KBr in the mixture is 0.1332 gm

Fraction of KBr  in mixture = 0.1332/ 0.560 = 0.2378

Q.4 27.6 gm of K₂CO₃  was treated by a series of reagents so as to convert all of its carbon to K₂Zn₃[ Fe(CN)₆]₂. Calculate the weight of the product.

Solution: - K₂ CO₃   → → → → K₂Zn₃[ Fe(CN)₆]₂

No. of moles of C atoms are conserved in the reaction

Apply POAC on C atoms

Moles of C atoms in K₂ CO₃   = Moles of C atoms in K₂Zn₃[ Fe(CN)₆]₂

1* no. of moles of  K₂ CO₃   =  12* no. of moles of  K₂Zn₃[ Fe(CN)₆]₂

Now you can calculate mass of K₂Zn₃[ Fe(CN)₆]₂

Hint:-

1* mass of K₂ CO₃  / m.wt  of K₂ CO₃  = 12* mass of  K₂Zn₃[ Fe(CN)₆]₂ /m.wt of K₂Zn₃[ Fe(CN)₆]

m.wt of  K₂Zn₃[ Fe(CN)₆]₂ = 698

m.wt of  K₂ CO₃   = 138

answer  mass of K₂Zn₃[ Fe(CN)₆]₂  =   11.6 gm