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