Chapter 3 : Fundamental concepts of Stoichiometry¶

Example 3.1 page no : 41¶

In [1]:
# Variables

m = 1000 * 0.4536;              #kg/min pounds
M = 30.24;                      #gm/mol avg. molecular weight

# Calculation
m1 = m * 60 / M;

# Result
print "molar folw rate = ",m1,"kmol/hr"

molar folw rate =  900.0 kmol/hr


Example 3.2 page no : 42¶

In [3]:
# Variables
MK = 39.1;         #potassium
MC = 12.0;         # carbon
MO = 16.;          # oxygen

# Calculation
MK2CO3 = MK * 2 + MC + MO * 3;
m = 691.;
N = m / MK2CO3;
A = 6.023 * 10**23;
molecules = N * A;

# Result
print "Total no. of molecules = %.4e molecules"%molecules

Total no. of molecules = 3.0115e+24 molecules


Example 3.3 page no : 43¶

In [5]:
# Variables
Na = 23.;                   #gm/mol  weight of sodium
MNa = 100.;                 #kg      sodium

# Calculation
N = MNa * 1000 / Na ;       #g-atoms
NNa2SO4 = N / 2;

# Result
print "(a) moles of sodium sulphate = %.3e mol"%NNa2SO4
MNa2SO4 = 142.06;
m = NNa2SO4 * MNa2SO4/1000;
print "(b)kilograms of sodium sulphate = %.2f kg"%m

(a) moles of sodium sulphate = 2.174e+03 mol
(b)kilograms of sodium sulphate = 308.83 kg


Example 3.4 page no : 43¶

In [6]:
# Variables
MFe = 55.85;
MO = 16.;
MS = 32.;
MFeS2 = MFe + MS * 2;        # molecular weight of FeS2
MFe2O3 = MFe * 2 + MO * 3;   # molecular weight of Fe2O3
MSO3 = MS + MO * 3;          # molecular weight of SO3
m1SO3 = 100.;                       #kg

# Calculation
N1 = m1SO3 / (MSO3);                #kmol
NFeS2 = N1 / 2;
mFeS2 = NFeS2 * MFeS2;

# Result
print "mass of pyrites to obtain 100kg of SO3 = %.2f kg"%mFeS2
m2SO3 = 50.;                        #kg
N2 = m2SO3 / (MSO3);                #kmol
NO2 = N2 * 15/8.;
mO2 = NO2 * MO * 2;
print "mass of Oxygen consumed to produce 50kg of SO3 = %.2f kg"%mO2

mass of pyrites to obtain 100kg of SO3 = 74.91 kg
mass of Oxygen consumed to produce 50kg of SO3 = 37.50 kg


Example 3.5 page no : 44¶

In [7]:
# Variables
MKClO3 = 122.55                 # weight of potassium
mKClO3 = 100.;                  #kg potassium

# Calculation
NKClO3 = mKClO3 / MKClO3;
NO2 = 3 * NKClO3 / 2;
V1 = 22.4143;                   #m**3/kmol;
V = V1 * NO2;

# Result
print "volume of oxygen produced = %.2f m**3"%V

volume of oxygen produced = 27.43 m**3


Example 3.6 page no : 44¶

In [9]:
# Variables
mH2 = 100.;                 #kg  hydrogen

# Calculation
NH2 = mH2/2.016;
NFe = 3. * NH2 / 4;
mFe = NFe * 55.85;

# Result
print "(a)mass of iron required = %.2f kg"%mFe
NH2O = NH2
mH2O = NH2O * 18;
print "mass of steam required = %.1f kg"%mH2O
V1 = 22.4143;               #m**3/kmol;
V = V1 * NH2;
print "(b)Volume of hydrogen = %.1f m**3"%V

# Answer may vary because of rounding error.

(a)mass of iron required = 2077.75 kg
mass of steam required = 892.9 kg
(b)Volume of hydrogen = 1111.8 m**3


Example 3.7 page no : 46¶

In [11]:
# variables
MCaCO3 = 100.08;    # molecular weight of CaCO3

# Calculation
GE = MCaCO3 / 2;

# Result
print "Gram equivalent wt. of CaCO3 =",GE,"g"

Gram equivalent wt. of CaCO3 = 50.04 g


Example 3.8 page no : 48¶

In [13]:
# variables
m1 = 1.;            #kg (mass in air)
m2 = 0.9;           #kg (mass in water)
m3 = 0.82;          #kg (mass in liquid)

# Calculation
L1 = m2 - m1;       #kg (loss of mass in water)
L2 = m3 - m1;       #kg (loss of mass in liquid)
sp_g = L2 /L1;

# Result
print "specific gravity of liquid = ",sp_g

specific gravity of liquid =  1.8


Example 3.9 page no : 49¶

In [15]:
# variables
m1 = 10.            #kg  liquid A
m2 = 5.             #kg  liquid B
sp_g1 = 1.17;       # gravity A
sp_g2 = 0.83;       # gravity B
Dwater = 1000.      #kg/m**3  water

# Calculation
DA = Dwater * sp_g1;
DB = Dwater * sp_g2;
V1 = m1 / DA;
V2 = m2 / DB;
V = V1 + V2;
Dmix = (m1 + m2)/ V ;
sp_g3 = Dmix

# Result
print "specific gravity of mixture = %.f kg/m**3"%sp_g3

specific gravity of mixture = 1029 kg/m**3


Example 3.10 page no : 49¶

In [17]:
# variables
Tw = 100.;              #Tw  baume scale

# Calculation
sp_g = Tw/200 + 1;
Be = 145 - 145/sp_g;

# Result
print "specific gravity on beume scale = %.1f Be"%Be

specific gravity on beume scale = 48.3 Be


Example 3.11 page no : 49¶

In [19]:
# variables
API1 = 30.                          #API gas oil
sp_g1 = 141.5/(131.5 + API1)        # (since, API = 141.5/sp_g -131.5)    gravity scale
Dwater = 999.                       #kg/m**3;  density of water

# Calculation
Doil1 = sp_g1 * Dwater;
V1 = 250.;                          #m**3
m1 = V1 * Doil1;
API2 = 15.;                         #API
sp_g2 = 141.5/(131.5 + API2);       # (since, API = 141.5/sp_g -131.5)
Dwater = 999.;                      #kg/m**3;
Doil2 = sp_g2 * Dwater;
V2 = 1000.;                         #m**3
m2 = V2 * Doil2;
Dmix = (m1 + m2)/(V1 + V2);

# Result
print "density of the mixture = %.f kg/m**3"%Dmix

density of the mixture = 947 kg/m**3


Example 3.12 page no : 51¶

In [20]:
# variables
m1 = 250.               #kg wet ammonium sulphate
mwater1 = 50.           #kg moisture

# Calculation
mdrysolid1 = m1 - mwater1;
wfe1 = mwater1 / m1;
wr1 = mwater1 / mdrysolid1;
wtpercentw1 = mwater1 * 100 / m1;
wtpercentd1 = mwater1 * 100 / mdrysolid1;
a = 90.                 #%
mwater2 = mwater1 * (1 - a/100);
m2 = mdrysolid1 + mwater2;
wfe2 = mwater2 / m2;
wr2 = mwater2 / mdrysolid1;
wtpercentw2 = mwater2 * 100 / m2;
wtpercentd2 = mwater2 * 100 / mdrysolid1;

# Result
print "(a)weight fraction of water at entrance =",wfe1
print "weight fraction of water at exit = %.3f"%wfe2
print "(b)weight ratio of water at entrance = ",wr1
print "weight ratio of water at exit = ",wr2
print "weight percent of moisture on wet basis at entrance = ",wtpercentw1,"%"
print "(c)weight percent of moisture on dry basis at entrance = ",wtpercentd1,"%"
print "weight percent of moisture on wet basis at exit = %.2f %%"%wtpercentw2
print "(d)weight percent of moisture on dry basis at exit = ",wtpercentd2,"%"

(a)weight fraction of water at entrance = 0.2
weight fraction of water at exit = 0.024
(b)weight ratio of water at entrance =  0.25
weight ratio of water at exit =  0.025
weight percent of moisture on wet basis at entrance =  20.0 %
(c)weight percent of moisture on dry basis at entrance =  25.0 %
weight percent of moisture on wet basis at exit = 2.44 %
(d)weight percent of moisture on dry basis at exit =  2.5 %


Example 3.13 page no : 53¶

In [21]:
# variables
mdrysolid = 100.                #kg  dry solids
percentin = 25.                 # water in the feed

# Calculation
mwaterin = mdrysolid * percentin / 100;
percentout = 2.5;
mwaterout = mdrysolid * percentout / 100;
mremoved = mwaterin - mwaterout;
percentremoved = mremoved *100 / mwaterin ;

# Result
print "percentage of water removed = ",percentremoved

percentage of water removed =  90.0


Example 3.14 page no : 53¶

In [23]:
# variables
m = 1.                      #kg  wet ammonium sulphate
percent1 = 20.              #%   water

# Calculation
mwaterin = m * percent1 / 100.
mdrysolid = m - mwaterin;
percent2 = 2.44;            #%
mout = mdrysolid / (1 - percent2/100);
mwaterout = mout - mdrysolid;
mremoved = mwaterin - mwaterout;
percentremoved = mremoved * 100 / mwaterin ;

# Result
print "weight of water removed = %.2f kg"%mremoved
print "percentage of water removed = %.f %%"%percentremoved

weight of water removed = 0.18 kg
percentage of water removed = 90 %


Example 3.15 page no : 55¶

In [1]:
# variables
mwater = 100.                #kg water
mNaCl = 35.8                 #kg solubility

# Calculation
msolu = mwater + mNaCl;
mfr = mNaCl / msolu;
mpr = mfr * 100;
MNaCl = 58.45;              #kg/kmol
NNaCl = mNaCl / MNaCl;
MH2O = 18.                  #kg/kmol
NH2O = mwater / MH2O;
Mfr = NNaCl / (NNaCl + NH2O);
Mpr = Mfr * 100;
N = NNaCl *1000 / mwater;

# Result
print "(a)mass fraction of NaCl = %.4f"%mfr
print "mass percent of NaCl= %.2f %%"%mpr
print "(b)mole fraction of NaCl = %.4f"%Mfr
print "mole percent of NaCl = %.2f %%"%Mpr
print "kmol NaCl per 1000 kg of water = %.3f kMol NaCl"%N

(a)mass fraction of NaCl = 0.2636
mass percent of NaCl= 26.36 %
(b)mole fraction of NaCl = 0.0993
mole percent of NaCl = 9.93 %
kmol NaCl per 1000 kg of water = 6.125 kMol NaCl


Example 3.16 page no : 55¶

In [26]:
# variables
Y = 0.015;                  #kg water vapour/kg dry air
Mair = 29.;                 #kg/kmol weight of air
Mwater = 18.016;            #kg/kmol

# Calculation
Nwater = Y / Mwater;        #kmol
Nair = 1 / Mair;            #kmol
Mpr = Nwater *100 / (Nwater + Nair);
Mr = Nwater / Nair;

# Result
print "(a)mole percent of water vapour = %.2f %%"%Mpr
print "(b) molal absolute humidity = %.4f kmol water/kmol dry air"%Mr

(a)mole percent of water vapour = 2.36 %
(b) molal absolute humidity = 0.0241 kmol water/kmol dry air


Example 3.17 page no : 56¶

In [27]:
# variables
msolu = 100.;                       #g basis
MK2CO3 = 138.20;                    #g/mol molecular weight
percent1 = 50.;                     #% mass of K2CO3

# Calculation
mK2CO3 = percent1 *msolu / 100;
NK2CO3 = mK2CO3 / MK2CO3;
mwater = msolu - mK2CO3;
Nwater = mwater / 18.06;
Mpr = NK2CO3 * 100 / (NK2CO3 + Nwater);
sp_gr  =1.53;
Vsolu = msolu/sp_gr;                #mL
Vwater = mwater / 1;                #mL
Vpr = Vwater * 100/ Vsolu;
Molality = NK2CO3 / (mwater * 10**-3);
Molarity = NK2CO3 / (Vsolu * 10**-3);
Eq_wt = MK2CO3 / 2;
No = mK2CO3/Eq_wt;
N = No / (Vsolu * 10**-3);

# Result
print "(a)Mole prcent of salt = %.2f %%"%Mpr
print "(b)Volume percent of water = %.2f %%"%Vpr
print "(c)Molality = %.3f mol/kg"%Molality
print "(d)Molarity = %.3f mol/L"%Molarity
print "(e)Normality = %.2f N"%N

(a)Mole prcent of salt = 11.56 %
(b)Volume percent of water = 76.50 %
(c)Molality = 7.236 mol/kg
(d)Molarity = 5.535 mol/L
(e)Normality = 11.07 N


Example 3.18 page no : 58¶

In [3]:
# variables
msolu = 100.;               #kg volume
percent1 = 60.;             #%  alcohol
Dwater = 998.;              #kg/m**3  water
Dalco = 798.;               #kg/m**3  alcohol
Dsolu = 895.;               #kg/m**3  solution

# Calculation
Vsolu = msolu/Dsolu;
malco = msolu * percent1 / 100;
Valco = malco / Dalco;
Vpr = Valco * 100 / Vsolu;
Malco = 46.048;             #kg/kmol
N = malco/Malco;
Molarity = N/(Vsolu );
mwater = msolu - malco;
Molality = N * 1000 /mwater;

# Result
print "(a)Volume percent of ethanol in solution = %.1f %%"%Vpr
print "(b)Molarity = %.2f mol/L"%Molarity
print "(c)Molality = %.3f mol/(kg of water)"% Molality

(a)Volume percent of ethanol in solution = 67.3 %
(b)Molarity = 11.66 mol/L
(c)Molality = 32.575 mol/(kg of water)


Example 3.19 page no : 63¶

In [29]:
#CO + CL2 = COCl2
import math
# Variables
Np = 12.;                   #moles phosgene
NCl2 = 3.;                  #moles chlorine
NCO = 8.;                   #moles carbon monoxide

# Calculation
N1Cl2 = NCl2 + Np;
N1CO = NCO + Np;
pr_ex = (N1CO - N1Cl2)* 100/N1Cl2;
pr_co = (N1Cl2-NCl2) * 100/ N1Cl2;
T = Np + NCl2 + NCO;
T1 = N1Cl2 + N1CO;
N = T / T1;

# Result
print "(a)percent excess of CO = %.2f %%"%pr_ex
print "(b)percent conversion = %.2f %%"%pr_co
print "(c)Moles of total products per mole of total reactants = %.3f"%N

(a)percent excess of CO = 33.33 %
(b)percent conversion = 80.00 %
(c)Moles of total products per mole of total reactants = 0.657


Example 3.21 page no : 64¶

In [1]:
# Variables

msolu = 100.;               #g  mold of feed mixture
MK2CO3 = 138.20;            #g/mol
ethylene = 53.89            # %
ethanol = 14.37             # %
ether = 1.8                 # %
water = 23.35               # %

# Calculation
Ethylene = ethylene * 83.57 / 100
Ethanol = ethanol*83.57 / 100
Ether = ether*83.57 / 100
Water = water*83.57 / 100
Inerts = 3                 # mol

conversion_ethylene = (60 - Ethylene)/60 * 100
yield_ethanol =  Ethanol/(60 - Ethylene)*100
yeild_ether = Ether*2/(60-Ethylene) * 100

# Result
print "Conversation of ethylene = %.2f %%"%conversion_ethylene
print "Yield of ethanol = %.2f %%"%yield_ethanol
print "Yield of Ether = %d %%"%yeild_ether

Conversation of ethylene = 24.94 %
Yield of ethanol = 80.25 %
Yield of Ether = 20 %