Chapter 25 :Stoichiometry

Example 25.2 , Page no:153

In [1]:
import math
from __future__ import division
 
#initialisation of variables
a=238.03; #atomic mass
m=75; #no. of moles

#CALCULATIONS
mass=m*a; #calculating mass of U
n=6.023*10**23; #avogadro's no.
no=m*n; #calculating no. of atoms

#RESULTS
print"Mass of U in gram =",round(mass,3);
print"No. of atoms =",'%.3E'%no;

Mass of U in gram = 17852.25
No. of atoms = 4.517E+25

Example 25.3 , Page no:153

In [2]:
import math
from __future__ import division
 
#initialisation of variables
a=63.54; #atomic mass of Cu
m=100; #mass of Cu

#CALCULATIONS
moles=m/a; #calculating moles of U
n=6.023*10**23; #avogadro's no.
no=moles*n; #calculating no. of atoms

#RESULTS
print"Maoles of U =",round(moles,3);
print"No. of atoms =",'%.3E'%no;

Maoles of U = 1.574
No. of atoms = 9.479E+23

Example 25.4 , Page no:153

In [3]:
import math
from __future__ import division
 
#initialisation of variables
no=10**24; #no of atoms
n=6.023*10**23; #avogadro's no.

#CALCULATIONS
moles=no/n; #calculating no. of moles

#RESULTS
print"Moles =",round(moles,3);

Moles = 1.66

Example 25.5 , Page no:154

In [4]:
import math
from __future__ import division
 
#initialisation of variables
c=12.01; #mass of carbon
h=1.008; #mass of hydrogen

#CALCULATIONS
mass=((2*c)+(4*h))*9.4; #calculating mass
n=6.023*10**23; #avogadro's no
ac=(2*9.4)*n; #calculating atoms of c

#RESULTS
print"Required mass =",round(mass,3);
print"Atoms of C =",'%.3E'%ac;

Required mass = 263.689
Atoms of C = 1.132E+25

Example 25.6 , Page no:154

In [5]:
import math
from __future__ import division
 
#initialisation of variables
c=12.01; #mass of carbon
h=1.008; #mass of hydrogen
o=16.00; #mass of oxygen

#CALCULATIONS
mass=((6*c)+(12*h)+(6*o)); #calculating formula mass
m=500*454; #mass of glucose in g
moles=m/mass; #moles 

#RESULTS
print"Moles =",round(moles,3);

Moles = 1260.019

Example 25.7 , Page no:155

In [6]:
import math
from __future__ import division
 
#initialisation of variables
pb=207.19; #mass of carbon
n=14.01; #mass of hydrogen
o=16.00; #mass of oxygen

#CALCULATIONS
mass=((1*pb)+(2*n)+(6*o)); #calculating formula mass
m=28.02; #no. of grams per mole
moles=m/mass; #moles 
moles1=moles*100;

#RESULTS
print"Proportion in percentage =",round(moles1,3);

Proportion in percentage = 8.46

Example 25.8 , Page no:155

In [7]:
import math
from __future__ import division
 
#initialisation of variables
m=50; #mass of N in g
a=22.99; #atomic mass in g/mole
ac=35.46; #atomic mass of chlorine
n=2.17; #no. of moles

#CALCULATIONS
mole=m/a; #moles of Na
mass=n*ac; #mass of Cl
ps=m/127; #proportion of sodium
pc=mass/127; #proportion of chlorine
ps1=ps*100;
pc1=pc*100;

#RESULTS
print"Moles of Na =",round(mole,3);
print"Mass oc Cl =",round(mass,3);
print"Proportion of Sodium =",round(ps1,3);
print"Proportion of Chlorine =",round(pc1,3);

Moles of Na = 2.175
Mass oc Cl = 76.948
Proportion of Sodium = 39.37
Proportion of Chlorine = 60.589

Example 25.9 , Page no:155

In [8]:
import math
from __future__ import division
 
#initialisation of variables
m=70; #mass in g
a=14.01; #atomic mass
h=1.008; #atomic mass of hydrogen

#CALCULATIONS
moles=m/a; #moles
mass=3*moles*h; #mass of H
ma=15+70; #mass of ammonia
mass1=(mass/ma)*100;
ma1=(m/ma)*100;

#RESULTS
print"Moles of N =",round(moles,3);
print"Mass of H =",round(mass,3);
print"Proportion of Hydrogen =",round(mass1,3);
print"Proportion of Nitrogen =",round(ma1,3);

Moles of N = 4.996
Mass of H = 15.109
Proportion of Hydrogen = 17.776
Proportion of Nitrogen = 82.353

Example 25.10 , Page no:155

In [9]:
import math
from __future__ import division
 
#initialisation of variables
m=200; #mass in g
o=16.00; #atomic mass
m1=6.25; #moles of S
as1=32.06;  #atomic mass of s

#CALCULATIONS
moles=m/o; #moles
m2=m1*as1;

#RESULTS
print"Mass of S =",round(m2,3);

Mass of S = 200.375

Example 25.11 , Page no:156

In [10]:
import math
from __future__ import division
 
#initialisation of variables
c=12.01; #mass of carbon
h=1.008; #mass of hydrogen
o=16.00; #mass of oxygen

#CALCULATIONS
mass=((2*c)+(2*h)); #calculating mass
moles=200/mass; #moles
mo=2*o*19.20; #mass of O2
mc=((2*o)+c)*15.36; #mass of CO2;

#RESULTS
print"Mass of O2 in g=",round(mo,3);
print"Mass of CO2 in g=",round(mc,3);

Mass of O2 in g= 614.4
Mass of CO2 in g= 675.994

Example 25.12 , Page no:156

In [11]:
import math
from __future__ import division
 
#initialisation of variables
m=1000; #mass of H2
fh=2.02; #formula mass of hydrogen
fo=32.00; #formula mass of oxygen

#CALCULATIONS
mass=62.5*18.02; #mass
M1=m/fh;
M2=m/fo;
M3=432.5*2.02;

#RESULTS
print"Moles of H2 =",round(M1,3);
print"MOles of O2 =",round(M2,3);
print"Mass of H2O =",round(mass,3);
print"Mass of H2 =",round(M3,3);

Moles of H2 = 495.05
MOles of O2 = 31.25
Mass of H2O = 1126.25
Mass of H2 = 873.65

Example 25.13 , Page no:156

In [12]:
import math
from __future__ import division
 
#initialisation of variables
Na=22.99; #mass of Na
S=32.06; #mass of S
O=16.00; #mass of O

#CALCULATIONS
mass=((2*Na)+(1*S)+(4*O)); #calculating mass
m=100/mass; #moles
m1=m*32.06;
m2=22.99*1.408;

#RESULTS
print"Moles =",round(m,3);
print"Mass of S =",round(m1,3);
print"Mass of Na =",round(m2,3);

Moles = 0.704
Mass of S = 22.571
Mass of Na = 32.37

Example 25.15 , Page no:157

In [13]:
import math
from __future__ import division
 
#initialisation of variables
a=128.8/32.06;
b=8.06/1.008;

#RESULTS
print"Moles of S =",round(a,3);
print"Moles of H =",round(b,3);

Moles of S = 4.017
Moles of H = 7.996

Example 25.16 , Page no:157

In [14]:
import math
from __future__ import division
 
#initialisation of variables
a=57.54/79.91;
b=17.29/12.01;
c=3.63/1.008;

#RESULTS
print"Moles of Br =",round(a,3);
print"Moles of C =",round(b,3);
print"Moles of H =",round(c,3);

Moles of Br = 0.72
Moles of C = 1.44
Moles of H = 3.601

Example 25.17 , Page no:157

In [15]:
import math
from __future__ import division
 
#initialisation of variables
a=100.9/12.01;
b=22.6/1.008;

#RESULTS
print"Moles of C =",round(a,3);
print"Moles of H =",round(b,3);

Moles of C = 8.401
Moles of H = 22.421