# Chapter 9 : The Chemical Reaction Equation and Stoichiometry¶

### Example 9.1 Page no. 228¶

In :
# variables
# Given
#Main eqn. C6H12O6 + aO2 ---> bCO2 + cH2O
# By carbon balance
b = 6 ;

#By hydrogen balance
c = 6;

# calculation
#Balancing oxygen in reaction
a = (c*1+b*2-6)/2.0;

#result
print 'Value of a is  %i'%a
print 'Value of b is  %i'%b
print 'Value of c is  %i'%c

Value of a is  6
Value of b is  6
Value of c is  6


### Example 9.2 Page no. 229¶

In :
# Variables
m_CO2 = 44.0 ;          #molecular wt-[g]
m_C7H16 = 100.1 ;       #molecular wt-[g]
p_con = 50. ;           # percentage conversion of CO2 to dry ice
amt_di = 500. ;         # amount of dry ice to be produce per hour-[kg]

# Calculation
# By using the given equation
amt_C7H16 = (amt_di*m_C7H16)/((p_con/100.)*m_CO2*7) ;# [kg]

# Result
print 'Amount of heptane required per hour to produce 500kg dry ice per hour is  %.1f kg.'%amt_C7H16

Amount of heptane required per hour to produce 500kg dry ice per hour is  325.0 kg.


### Example 9.3 Page no. 230¶

In :
# Variables
m_CaCO3 = 100.1 ;              #molecular wt-[g]
m_MgCO3 = 84.32 ;              #molecular wt-[g]
m_CaO = 56.08 ;                #molecular wt-[g]
m_MgO = 40.32 ;                #molecular wt-[g]
m_CO2 = 44.0 ;                 #molecular wt-[g]

# Limestone analysis
p_CaCO3 = 92.89 ;              # percentage of CaCO3
p_MgCO3 = 5.41 ;               #  percentage of MgCO3
inrt = 1.7 ;                   #percentage of inert

# Calculation and Results
#(a)
amt_CaO  = (((p_CaCO3/100)*m_CaO)/m_CaCO3)*2000 ;      #Pounds of CaO produced from 1 ton(2000lb) of limestone
print ' Amount of CaO produced from 1 ton(2000lb) of limestone is  %.0f lb.'%amt_CaO

#(b)
mol_CaCO3 = (p_CaCO3/100)/m_CaCO3 ;      # lb mol of CaCO3
mol_MgCO3 = (p_MgCO3/100)/m_MgCO3 ;      # lb mol of MgCO3
total_mol = mol_CaCO3+mol_MgCO3;
amt_CO2 = total_mol*m_CO2 ;              # Amount of CO2 recovered per pound of limestone-[lb]
print '  Amount of CO2 recovered per pound of limestone is  %.3f lb.'%amt_CO2

#(c)
amt_CaO = m_CaO*mol_CaCO3 ;              # since lb mol of CaO  =  CaCO3
amt_MgO = m_MgO*mol_MgCO3 ;              # since lb mol of MgO  =  MgCO3
total_lime =  amt_CaO+amt_MgO+(inrt)/100 ;     # total amount of lime per pound limestone
amt_lmst = 2000/total_lime ;              # Amount of limestone required to make 1 ton(2000lb) of lime
print '  Amount of limestone required to make 1 ton(2000lb) of lime   %.1f lb.'%amt_lmst

 Amount of CaO produced from 1 ton(2000lb) of limestone is  1041 lb.
Amount of CO2 recovered per pound of limestone is  0.437 lb.
Amount of limestone required to make 1 ton(2000lb) of lime   3550.7 lb.


### Example 9.4 Page no. 235¶

In :
# Variables
f_NH3 = 5. ;        # NH3 in feed-[g]
f_N2 =  100. ;      # N2 in feed-[g]
f_H2 =  50. ;       # H2 in feed-[g]
p_NH3 = 90.         # NH3 in product-[g]
m_NH3  = 17. ;      # Molecular wt. of NH3-[g]
m_N2  = 28. ;       # Molecular wt. of N2-[g]
m_H2  = 2. ;        # Molecular wt. of H2-[g]

# Calculations
# Extent of reaction can be calculated by using eqn. 9.3
# For NH3
ni = p_NH3/m_NH3 ;       #[g mol NH3]
nio = f_NH3/m_NH3 ;      #[g mol NH3]
vi = 2 ;                 # coefficint of NH3
ex_r =  (ni-nio)/vi      # Extent of reaction - moles reacting

#Determine H2 and N2 in product of reaction by Eqn. 9.4
# For N2
nio_N2 = f_N2/m_N2 ;     #[g mol N2]
vi_N2 = -1 ;             # coefficint of N2
ni_N2 = nio_N2 + vi_N2*ex_r ;      #N2 in product of reaction-[g moles ]
m_N2 = ni_N2*m_N2 ;                # mass of N2 in product of reaction-[g]

# Results
print '  N2 in product of reaction is  %.2f g moles  '%ni_N2
print '   Mass of N2 in product of reaction is  %.2f g   '%m_N2
# For H2
nio_H2 = f_H2/m_H2 ;               #[g mol H2]
vi_H2 = -3 ;                       # coefficint of H2
ni_H2 = nio_H2 + vi_H2*ex_r ;      #H2 in product of reaction-[g moles ]
m_H2 = ni_H2*m_H2 ;                # mass of H2 in product of reaction-[g]
print '   H2 in product of reaction is  %.2f g moles  '%ni_H2
print '  Mass of H2 in product of reaction is  %.2f g   '%m_H2

# ARP
m_SO2 = 64. ;                      # Molecular wt.of SO2-[g]
mol_SO2 =  2. ;                    # moles of SO2
ARP = (1./m_NH3)/(mol_SO2/m_SO2);
print '  ARP is  %.2f    '%ARP

  N2 in product of reaction is  1.07 g moles
Mass of N2 in product of reaction is  30.00 g
H2 in product of reaction is  17.50 g moles
Mass of H2 in product of reaction is  35.00 g
ARP is  1.88


### Example 9.5 Page no. 238¶

In :
# Variables
f_N2 = 10. ;          # N2 in feed-[g]
f_H2 = 10. ;          # H2 in feed-[g]
m_NH3 = 17.02         # Molecular wt. of NH3-[g]
m_N2 = 28. ;          # Molecular wt. of N2-[g]
m_H2 = 2. ;           # Molecular wt. of H2-[g]

# Calculations
# Extent of reaction can be calculated by using eqn. 9.3
# Based on N2
nio_N2 = f_N2/m_N2         #[g mol N2]
vi_N2 = -1 ;               # coefficint of N2
ex_N2 = -(nio_N2)/vi_N2 ;  # Max. extent of reaction based on N2

# Based on H2
nio_H2 = f_H2/m_H2 ;       #[g mol H2]
vi_H2 = -3 ;               # coefficint of H2
ex_H2 = -(nio_H2)/vi_H2    # Max. extent of reaction based on H2

#(a)
vi_NH3 = 2 ;               # coefficint of NH3
mx_NH3 = ex_N2*vi_NH3*m_NH3 ;    # Max. amount of NH3 that can be produced

# Results
print ' (a) Max. amount of NH3 that can be produced is %.1f g'%mx_NH3

#(b) and (c)
if (ex_H2 > ex_N2 ):
print '  (b) N2 is limiting reactant  '
print '  (c) H2 is excess reactant  '
ex_r = ex_N2
else:
print '  (b) H2 is limiting reactant  '
print '  (c) N2 is excess reactant  '
ex_r = ex_H2 ;

 (a) Max. amount of NH3 that can be produced is 12.2 g
(b) N2 is limiting reactant
(c) H2 is excess reactant


### Example 9.6 Page no. 242¶

In :
# variables
#(a)
mol_bms =  0.59 ;      # Biomass produced per g mol of glucose-[g mol biomass/ g mol glucose]
mw_bms =  23.74 ;      # molecular wt. of biomass -[g]
mw_gls =  180.0 ;      # molecular wt. of glucose -[g]

# calculations and Results
ms_bms =  (mol_bms*mw_bms)/mw_gls ;     # Biomass produced per gram of glucose-[g biomass/ g glucose]
print '(a) Biomass produced per gram of glucose is %.4f g biomass/ g glucose.'%ms_bms

#(b)
mol_etol = 1.3 ;       #Ethanol produced per g mol of glucose-[g mol ethanol/ g mol glucose]
mw_etol =  46.0 ;      # molecular wt. of ethanol -[g]
ms_etol =  (mol_etol*mw_etol)/mw_gls ;     # Ethanol produced per gram of glucose-[g ethanol/ g glucose]
print ' (b) Ethanol produced per gram of glucose is %.3f g ethanol/ g glucose.'%ms_etol

(a) Biomass produced per gram of glucose is 0.0778 g biomass/ g glucose.
(b) Ethanol produced per gram of glucose is 0.332 g ethanol/ g glucose.


### Example 9.7 Page no. 243¶

In :
# Variables
# By using reaction (a)
H2_a = 3-0.50        # H2 produced in reaction (a)

# Calculations
C_a = (2./3)*H2_a ;  # Nanotubes(the C) produced by reaction (a)
sel = C_a/0.50 ;     # Selectivity of C reletive to C2H4-[g mol C/ g mol C2H4]

# Results
print 'Selectivity of C reletive to C2H4 is %.2f g mol C/ g mol C2H4.'%sel

Selectivity of C reletive to C2H4 is 3.33 g mol C/ g mol C2H4.


### Example 9.8 Page no. 244¶

In :
# Variables
m_C3H6 = 42.08          # molecular wt. of propene-[g]
m_C3H5Cl = 76.53 ;      # molecular wt. of C3H5Cl-[g]
m_C3H6Cl2 = 112.99 ;    # molecular wt. of C3H6Cl2-[g]

# Product analysis
pml_Cl2 = 141.0 ;       # [g mol]
pml_C3H6 = 651.0 ;      #[g mol]
pml_C3H5Cl = 4.6 ;      # [g mol]
pml_C3H6Cl2 = 24.5 ;    # [g mol]
pml_HCL = 4.6 ;         #[g mol]

# Calculation & Results
#(a)
a_Cl = pml_C3H5Cl;      # Chlorine reacted by eqn.(a)
b_Cl = pml_C3H6Cl2 ;    # Chlorine reacted by eqn.(b)
fed_Cl = pml_Cl2+a_Cl+b_Cl ;    # Total chlorine fed to reactor-[g mol]

#by analysing reaction (a) and (b)
a_C3H6 = a_Cl+b_Cl ;     # C3H6 reacted by reaction (a)
fed_C3H6 = pml_C3H6+a_C3H6 ;      #Total C3H6 fed to reactor-[g mol]

print '(a) Total chlorine fed to reactor is %.2f  g mol '%fed_Cl
print '     Total C3H6 fed to reactor is %.2f  g mol '%fed_C3H6

#(b) and (c)
# Extent of reaction can be calculated by using eqn. 9.3
# Based on C3H6
nio_C3H6 = fed_C3H6 ;       #[g mol C3H6]
vi_C3H6 = -1 ;              # coefficint of C3H6
ex_C3H6 = -(nio_C3H6)/vi_C3H6 ;     # Max. extent of reaction based on C3H6

# Based on Cl2
nio_Cl2 =  fed_Cl;          #[g mol Cl2]
vi_Cl2 = -1 ;               # coefficint of Cl2
ex_Cl2 = -(nio_Cl2)/vi_Cl2 ;# Max. extent of reaction based on Cl2

if (ex_Cl2 > ex_C3H6 ):
print '  (b) C3H6 is limiting reactant  '
print '    (c)Cl2 is excess reactant  '
ex_r = ex_C3H6;
else:
print '  (b) Cl2 is limiting reactant  '
print ' (c) C3H6 is excess reactant  '
ex_r = ex_Cl2;

#(d)
fr_cn = pml_C3H5Cl/fed_C3H6 ;      #Fractional conversion of C3H6 to C3H5Cl
print '  (d) Fractional conversion of C3H6 to C3H5Cl is %.2e  '%fr_cn

#(e)
sel = pml_C3H5Cl/pml_C3H6Cl2 ;     # Selectivity of C3H5Cl relative to C3H6Cl2
print '  (e) Selectivity of C3H5Cl relative to C3H6Cl2 is %.2f g mol C3H5Cl/g mol C3H6Cl2  '%sel

#(f)
yld = (m_C3H5Cl*pml_C3H5Cl)/(m_C3H6*fed_C3H6)   # Yield of C3H5Cl per g C3H6 fed to reactor
print '  (f) Yield of C3H5Cl per g C3H6 fed to reactor is %.3f g C3H5Cl/g C3H6  '%yld

#(g)
vi_C3H5Cl = 1  ;                 # coefficint of C3H5Cl
vi_C3H6Cl2 = 1  ;                # coefficint of C3H6Cl2
ex_a = (pml_C3H5Cl-0)/vi_C3H5Cl ;      # Extent of reaction a as C3H5Cl is produced only in reaction a
ex_b = (pml_C3H6Cl2-0)/vi_C3H6Cl2 ;    # Extent of reaction b as C3H6Cl2 is produced only in reaction b
print '  (g) Extent of reaction a as C3H5Cl is produced only in reaction a is %.1f   '%ex_a
print '     Extent of reaction b as C3H6Cl2 is produced only in reaction b %.1f   '%ex_b

#(h)
in_Cl = fed_Cl*2 ;        #Entering Cl -[g mol]
out_Cl  = pml_HCL ;       # Exiting Cl in HCl-[g mol]
ef_w = out_Cl/in_Cl ;     # Mole efficiency of waste
ef_pr =  1-ef_w ;         # Mole efficiency of product
print ' (h) Mole efficiency of product is %.3f '%ef_pr

(a) Total chlorine fed to reactor is 170.10  g mol
Total C3H6 fed to reactor is 680.10  g mol
(b) Cl2 is limiting reactant
(c) C3H6 is excess reactant
(d) Fractional conversion of C3H6 to C3H5Cl is 6.76e-03
(e) Selectivity of C3H5Cl relative to C3H6Cl2 is 0.19 g mol C3H5Cl/g mol C3H6Cl2
(f) Yield of C3H5Cl per g C3H6 fed to reactor is 0.012 g C3H5Cl/g C3H6
(g) Extent of reaction a as C3H5Cl is produced only in reaction a is 4.6
Extent of reaction b as C3H6Cl2 is produced only in reaction b 24.5
(h) Mole efficiency of product is 0.986

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