Download Stoichiometry - Shailendra Kumar Chemistry

PHYSICAL CHEMISTRY
By: Shailendra Kumar
PHYSICAL CHEMISTRY
by: SHAILENDRA KR.
Meq. Approach
Classes at: SCIENCE TUTORIALS; Opp. Khuda Baksh Library, Ashok Rajpath, Patna
PIN POINT STUDY CIRCLE; House No. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Problem Solving Method For Stoichiometry
1.
1. Principle of atom conservation (POAC)
2.
Millimol method
3. Factor level method
4.
Milliequivalent method
Principle of atom conservation (POAC): Principle of atom conservation based on the conservation of
mass of atom or mole of atom, or number of atom. Balanced equations are not needed for solving
stoichiometry problem by POAC method.
Consider an example (a):
KCl + O2 (No need of balanced equation)
KClO3
Conservation of K atom:
Total mol of K atom in KClO3 = Total mol of K atom in KCl
1 × no of mol of KClO3 = 1 × no of mol of KCl
Conservation of Cl atom:
Total mol of Cl atom in KClO3 = Total mol of Cl atom in KCl.
1 × no of mol of KClO3 = 1 no of mol of KCl
Conservation of oxygen atom:
Total mol of oxygen atom in KClO3 = Total mol of O atom in O2
3 × no of mol of KClO3 = 2 × no of mol of O2
Example (b):
O3
O2 (No need of balanced equation)
Total mol of oxygen atom in O3 = Total mol of oxygen atom in O2
3 × no of mol of O3 = 2 × no of mol of O2
Prob.1:
Calculate weight of O2 produced when 12.25 gm KClO3 heated according to the equation
KClO3
KCl + O2
Solution: In this problem we must derive the relation between KClO3 and O2.
Conservation of O atom:
3 × no of mol of KClO3 = 2 × no of mol of O2
3×
12.25
wt of O2
12.25
32
16
weight of O2 = 4.8 gm.
=2×
Prob.2:
In the above problem calculate volume of O2 evolved at S.T.P.
Solution: 3 × 12.25 = 2 × Volume of O2
Volume of O2 = 3.36 liter
122.5
22.4
10
11.2
1.
2.
Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 1
PHYSICAL CHEMISTRY
By: Shailendra Kumar
Problem for practice:
(1)
27.6 gm of K2CO3 was treated by a series of reagents so as to convert all its carbon to K2Zn3[Fe(CN)6]2.
Calculate the weight of the product.
Ans:
(2)
11.6 gm.
A sample of KClO3 on decomposition yielded 448 ml of oxygen gas at NTP. Calculate
(i) weight of oxygen produced,
(ii) weight of KClO3 originally taken and
(iii) weight of KCl produced
Ans: (i) 0.64 gm
(3)
(ii) 1.634 gm
(iii) 0.9937 gm
What weight of Co is required to form Re2(Co)10 from 2.50 gm of Re2O7 according to the unbalanced
reaction ?
Re2O7
+ Co
Re2(Co)10 + Co2
Re = 186.2
C = 12 O = 16
Ans: wt of Co = 2.46 gm.
2.
Millimol method:
2A + 3B
4C + 5D (Need of balanced equation)
Mol of A reacted
=
Mol of B reacted
=
3
2
Mol of C fromed
4
=
Mol of D fromed
5
This relation is derived by rule of chemical kinetics .
+ ∆[A]
– ∆[A]
– ∆[B]
+ ∆[A]
=
=
=
∆t
4∆
∆t
∆2∆
∆t
3∆
∆t
5∆
– d[B]
+ d[C]
+ d[D]
– d[A]
Instantaneous rate of reaction =
=
=
=
3dt
4dt
5dt
2dt
Average rate of reaction =
–∆
∆[A] = mol of A Reacted
–∆
∆[A] = mol of A Reacted
+∆
∆[A] = mol of A Reacted
+∆
∆[A] = mol of A Reacted
2O3
=
(
3O2
Mol of CO3
2
Mol of O2
=
Millimol of A
=
2
(MV)A
2
( Same relation obtain by POAC method)
3
Mol of B
Mol of A
=
3
2
=
1.
2.
=
=
=
Mol of D
Mol of C
=
) × 1000
5
4
Millimol of B
=
3
(MV)B
3
=
Millimol of C
=
4
(MV)C
4
=
Millimol of D
5
(MV)D
5
Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 2
PHYSICAL CHEMISTRY
[Q]
By: Shailendra Kumar
Mol of A means mol of A reacted
Mol of B
,,
,,
B reacted
,,
,, C
,,
,,
C formed
,, ,, D
,,
,,
D formed
Calculate volume of 0.5 M NaOH needed for complete reaction with 100 ml of 0.10 M H2SO4.
2NaOH + H2SO4
Na2SO4 + 2H2O
Millimol of H2SO4
1
Millimol of NaOH
=
2
0.5 × V(ml)
2
V(ml)
[Q]
100 × 0.10
1
=
10 × 2
0.5
=
= 40 ml.
Calculate wt of Na2S2O3 required for complete reaction with 100 ml of 0.5 M I2 according to the
equation.
2Na2S2O3 + I2
Na2S4O6 + 2NaI
Millimol of Na2S2O3
=
Millimol of I2
1
2
Millimol of Na2S2O3
158 × 2
× 1000 =
100 × 0.5
1
Wt of Na2S2O3 = 100 × 0.5 × 158 × 2
15800 gm
1000
1.
2.
= 15.8 gm
Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 3