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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 9 : The Chemical Reaction Equation and Stoichiometry"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.1 Page no. 228\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# variables\n",
"# Given \n",
"#Main eqn. C6H12O6 + aO2 ---> bCO2 + cH2O\n",
"# By carbon balance\n",
"b = 6 ;\n",
"\n",
"#By hydrogen balance\n",
"c = 6;\n",
"\n",
"# calculation\n",
"#Balancing oxygen in reaction\n",
"a = (c*1+b*2-6)/2.0;\n",
"\n",
"#result\n",
"print 'Value of a is %i'%a\n",
"print 'Value of b is %i'%b\n",
"print 'Value of c is %i'%c"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Value of a is 6\n",
"Value of b is 6\n",
"Value of c is 6\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 9.2 Page no. 229\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"m_CO2 = 44.0 ; #molecular wt-[g]\n",
"m_C7H16 = 100.1 ; #molecular wt-[g]\n",
"p_con = 50. ; # percentage conversion of CO2 to dry ice\n",
"amt_di = 500. ; # amount of dry ice to be produce per hour-[kg]\n",
"\n",
"# Calculation\n",
"# By using the given equation \n",
"amt_C7H16 = (amt_di*m_C7H16)/((p_con/100.)*m_CO2*7) ;# [kg]\n",
"\n",
"# Result\n",
"print 'Amount of heptane required per hour to produce 500kg dry ice per hour is %.1f kg.'%amt_C7H16"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Amount of heptane required per hour to produce 500kg dry ice per hour is 325.0 kg.\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 9.3 Page no. 230\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"m_CaCO3 = 100.1 ; #molecular wt-[g]\n",
"m_MgCO3 = 84.32 ; #molecular wt-[g]\n",
"m_CaO = 56.08 ; #molecular wt-[g]\n",
"m_MgO = 40.32 ; #molecular wt-[g]\n",
"m_CO2 = 44.0 ; #molecular wt-[g]\n",
"\n",
"# Limestone analysis\n",
"p_CaCO3 = 92.89 ; # percentage of CaCO3\n",
"p_MgCO3 = 5.41 ; # percentage of MgCO3 \n",
"inrt = 1.7 ; #percentage of inert\n",
"\n",
"# Calculation and Results\n",
"#(a)\n",
"amt_CaO = (((p_CaCO3/100)*m_CaO)/m_CaCO3)*2000 ; #Pounds of CaO produced from 1 ton(2000lb) of limestone\n",
"print ' Amount of CaO produced from 1 ton(2000lb) of limestone is %.0f lb.'%amt_CaO\n",
"\n",
"#(b)\n",
"mol_CaCO3 = (p_CaCO3/100)/m_CaCO3 ; # lb mol of CaCO3\n",
"mol_MgCO3 = (p_MgCO3/100)/m_MgCO3 ; # lb mol of MgCO3\n",
"total_mol = mol_CaCO3+mol_MgCO3;\n",
"amt_CO2 = total_mol*m_CO2 ; # Amount of CO2 recovered per pound of limestone-[lb]\n",
"print ' Amount of CO2 recovered per pound of limestone is %.3f lb.'%amt_CO2\n",
"\n",
"#(c)\n",
"amt_CaO = m_CaO*mol_CaCO3 ; # since lb mol of CaO = CaCO3\n",
"amt_MgO = m_MgO*mol_MgCO3 ; # since lb mol of MgO = MgCO3\n",
"total_lime = amt_CaO+amt_MgO+(inrt)/100 ; # total amount of lime per pound limestone\n",
"amt_lmst = 2000/total_lime ; # Amount of limestone required to make 1 ton(2000lb) of lime \n",
"print ' Amount of limestone required to make 1 ton(2000lb) of lime %.1f lb.'%amt_lmst"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Amount of CaO produced from 1 ton(2000lb) of limestone is 1041 lb.\n",
" Amount of CO2 recovered per pound of limestone is 0.437 lb.\n",
" Amount of limestone required to make 1 ton(2000lb) of lime 3550.7 lb.\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 9.4 Page no. 235\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"f_NH3 = 5. ; # NH3 in feed-[g]\n",
"f_N2 = 100. ; # N2 in feed-[g]\n",
"f_H2 = 50. ; # H2 in feed-[g]\n",
"p_NH3 = 90. # NH3 in product-[g]\n",
"m_NH3 = 17. ; # Molecular wt. of NH3-[g]\n",
"m_N2 = 28. ; # Molecular wt. of N2-[g]\n",
"m_H2 = 2. ; # Molecular wt. of H2-[g]\n",
"\n",
"# Calculations \n",
"# Extent of reaction can be calculated by using eqn. 9.3 \n",
"# For NH3\n",
"ni = p_NH3/m_NH3 ; #[g mol NH3]\n",
"nio = f_NH3/m_NH3 ; #[g mol NH3]\n",
"vi = 2 ; # coefficint of NH3\n",
"ex_r = (ni-nio)/vi # Extent of reaction - moles reacting\n",
"\n",
"#Determine H2 and N2 in product of reaction by Eqn. 9.4\n",
"# For N2\n",
"nio_N2 = f_N2/m_N2 ; #[g mol N2]\n",
"vi_N2 = -1 ; # coefficint of N2\n",
"ni_N2 = nio_N2 + vi_N2*ex_r ; #N2 in product of reaction-[g moles ]\n",
"m_N2 = ni_N2*m_N2 ; # mass of N2 in product of reaction-[g]\n",
"\n",
"# Results\n",
"print ' N2 in product of reaction is %.2f g moles '%ni_N2\n",
"print ' Mass of N2 in product of reaction is %.2f g '%m_N2\n",
"# For H2\n",
"nio_H2 = f_H2/m_H2 ; #[g mol H2]\n",
"vi_H2 = -3 ; # coefficint of H2\n",
"ni_H2 = nio_H2 + vi_H2*ex_r ; #H2 in product of reaction-[g moles ]\n",
"m_H2 = ni_H2*m_H2 ; # mass of H2 in product of reaction-[g]\n",
"print ' H2 in product of reaction is %.2f g moles '%ni_H2\n",
"print ' Mass of H2 in product of reaction is %.2f g '%m_H2\n",
"\n",
"# ARP\n",
"m_SO2 = 64. ; # Molecular wt.of SO2-[g]\n",
"mol_SO2 = 2. ; # moles of SO2\n",
"ARP = (1./m_NH3)/(mol_SO2/m_SO2);\n",
"print ' ARP is %.2f '%ARP"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" N2 in product of reaction is 1.07 g moles \n",
" Mass of N2 in product of reaction is 30.00 g \n",
" H2 in product of reaction is 17.50 g moles \n",
" Mass of H2 in product of reaction is 35.00 g \n",
" ARP is 1.88 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 9.5 Page no. 238\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"f_N2 = 10. ; # N2 in feed-[g]\n",
"f_H2 = 10. ; # H2 in feed-[g]\n",
"m_NH3 = 17.02 # Molecular wt. of NH3-[g]\n",
"m_N2 = 28. ; # Molecular wt. of N2-[g]\n",
"m_H2 = 2. ; # Molecular wt. of H2-[g]\n",
"\n",
"# Calculations\n",
"# Extent of reaction can be calculated by using eqn. 9.3 \n",
"# Based on N2\n",
"nio_N2 = f_N2/m_N2 #[g mol N2]\n",
"vi_N2 = -1 ; # coefficint of N2\n",
"ex_N2 = -(nio_N2)/vi_N2 ; # Max. extent of reaction based on N2\n",
"\n",
"# Based on H2\n",
"nio_H2 = f_H2/m_H2 ; #[g mol H2]\n",
"vi_H2 = -3 ; # coefficint of H2\n",
"ex_H2 = -(nio_H2)/vi_H2 # Max. extent of reaction based on H2\n",
"\n",
"#(a)\n",
"vi_NH3 = 2 ; # coefficint of NH3\n",
"mx_NH3 = ex_N2*vi_NH3*m_NH3 ; # Max. amount of NH3 that can be produced\n",
"\n",
"# Results\n",
"print ' (a) Max. amount of NH3 that can be produced is %.1f g'%mx_NH3\n",
"\n",
"#(b) and (c)\n",
"if (ex_H2 > ex_N2 ):\n",
" print ' (b) N2 is limiting reactant '\n",
" print ' (c) H2 is excess reactant '\n",
" ex_r = ex_N2\n",
"else:\n",
" print ' (b) H2 is limiting reactant '\n",
" print ' (c) N2 is excess reactant '\n",
" ex_r = ex_H2 ;"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" (a) Max. amount of NH3 that can be produced is 12.2 g\n",
" (b) N2 is limiting reactant \n",
" (c) H2 is excess reactant \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 9.6 Page no. 242\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# variables\n",
"#(a)\n",
"mol_bms = 0.59 ; # Biomass produced per g mol of glucose-[g mol biomass/ g mol glucose]\n",
"mw_bms = 23.74 ; # molecular wt. of biomass -[g]\n",
"mw_gls = 180.0 ; # molecular wt. of glucose -[g]\n",
"\n",
"# calculations and Results\n",
"ms_bms = (mol_bms*mw_bms)/mw_gls ; # Biomass produced per gram of glucose-[g biomass/ g glucose]\n",
"print '(a) Biomass produced per gram of glucose is %.4f g biomass/ g glucose.'%ms_bms\n",
"\n",
"#(b)\n",
"mol_etol = 1.3 ; #Ethanol produced per g mol of glucose-[g mol ethanol/ g mol glucose]\n",
"mw_etol = 46.0 ; # molecular wt. of ethanol -[g]\n",
"ms_etol = (mol_etol*mw_etol)/mw_gls ; # Ethanol produced per gram of glucose-[g ethanol/ g glucose]\n",
"print ' (b) Ethanol produced per gram of glucose is %.3f g ethanol/ g glucose.'%ms_etol"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Biomass produced per gram of glucose is 0.0778 g biomass/ g glucose.\n",
" (b) Ethanol produced per gram of glucose is 0.332 g ethanol/ g glucose.\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 9.7 Page no. 243\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"# By using reaction (a)\n",
"H2_a = 3-0.50 # H2 produced in reaction (a)\n",
"\n",
"# Calculations\n",
"C_a = (2./3)*H2_a ; # Nanotubes(the C) produced by reaction (a)\n",
"sel = C_a/0.50 ; # Selectivity of C reletive to C2H4-[g mol C/ g mol C2H4]\n",
"\n",
"# Results\n",
"print 'Selectivity of C reletive to C2H4 is %.2f g mol C/ g mol C2H4.'%sel"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Selectivity of C reletive to C2H4 is 3.33 g mol C/ g mol C2H4.\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 9.8 Page no. 244\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"m_C3H6 = 42.08 # molecular wt. of propene-[g]\n",
"m_C3H5Cl = 76.53 ; # molecular wt. of C3H5Cl-[g]\n",
"m_C3H6Cl2 = 112.99 ; # molecular wt. of C3H6Cl2-[g]\n",
"\n",
"# Product analysis\n",
"pml_Cl2 = 141.0 ; # [g mol]\n",
"pml_C3H6 = 651.0 ; #[g mol]\n",
"pml_C3H5Cl = 4.6 ; # [g mol]\n",
"pml_C3H6Cl2 = 24.5 ; # [g mol]\n",
"pml_HCL = 4.6 ; #[g mol]\n",
"\n",
"# Calculation & Results\n",
"#(a)\n",
"a_Cl = pml_C3H5Cl; # Chlorine reacted by eqn.(a)\n",
"b_Cl = pml_C3H6Cl2 ; # Chlorine reacted by eqn.(b)\n",
"fed_Cl = pml_Cl2+a_Cl+b_Cl ; # Total chlorine fed to reactor-[g mol]\n",
"\n",
"#by analysing reaction (a) and (b)\n",
"a_C3H6 = a_Cl+b_Cl ; # C3H6 reacted by reaction (a)\n",
"fed_C3H6 = pml_C3H6+a_C3H6 ; #Total C3H6 fed to reactor-[g mol]\n",
"\n",
"\n",
"print '(a) Total chlorine fed to reactor is %.2f g mol '%fed_Cl\n",
"print ' Total C3H6 fed to reactor is %.2f g mol '%fed_C3H6\n",
"\n",
"#(b) and (c)\n",
"# Extent of reaction can be calculated by using eqn. 9.3 \n",
"# Based on C3H6\n",
"nio_C3H6 = fed_C3H6 ; #[g mol C3H6]\n",
"vi_C3H6 = -1 ; # coefficint of C3H6\n",
"ex_C3H6 = -(nio_C3H6)/vi_C3H6 ; # Max. extent of reaction based on C3H6\n",
"\n",
"# Based on Cl2\n",
"nio_Cl2 = fed_Cl; #[g mol Cl2]\n",
"vi_Cl2 = -1 ; # coefficint of Cl2\n",
"ex_Cl2 = -(nio_Cl2)/vi_Cl2 ;# Max. extent of reaction based on Cl2\n",
"\n",
"if (ex_Cl2 > ex_C3H6 ):\n",
" print ' (b) C3H6 is limiting reactant '\n",
" print ' (c)Cl2 is excess reactant '\n",
" ex_r = ex_C3H6;\n",
"else:\n",
" print ' (b) Cl2 is limiting reactant '\n",
" print ' (c) C3H6 is excess reactant '\n",
" ex_r = ex_Cl2;\n",
"\n",
"#(d)\n",
"fr_cn = pml_C3H5Cl/fed_C3H6 ; #Fractional conversion of C3H6 to C3H5Cl\n",
"print ' (d) Fractional conversion of C3H6 to C3H5Cl is %.2e '%fr_cn\n",
"\n",
"#(e)\n",
"sel = pml_C3H5Cl/pml_C3H6Cl2 ; # Selectivity of C3H5Cl relative to C3H6Cl2\n",
"print ' (e) Selectivity of C3H5Cl relative to C3H6Cl2 is %.2f g mol C3H5Cl/g mol C3H6Cl2 '%sel\n",
"\n",
"#(f)\n",
"yld = (m_C3H5Cl*pml_C3H5Cl)/(m_C3H6*fed_C3H6) # Yield of C3H5Cl per g C3H6 fed to reactor\n",
"print ' (f) Yield of C3H5Cl per g C3H6 fed to reactor is %.3f g C3H5Cl/g C3H6 '%yld\n",
"\n",
"#(g)\n",
"vi_C3H5Cl = 1 ; # coefficint of C3H5Cl\n",
"vi_C3H6Cl2 = 1 ; # coefficint of C3H6Cl2\n",
"ex_a = (pml_C3H5Cl-0)/vi_C3H5Cl ; # Extent of reaction a as C3H5Cl is produced only in reaction a\n",
"ex_b = (pml_C3H6Cl2-0)/vi_C3H6Cl2 ; # Extent of reaction b as C3H6Cl2 is produced only in reaction b\n",
"print ' (g) Extent of reaction a as C3H5Cl is produced only in reaction a is %.1f '%ex_a\n",
"print ' Extent of reaction b as C3H6Cl2 is produced only in reaction b %.1f '%ex_b\n",
"\n",
"#(h)\n",
"in_Cl = fed_Cl*2 ; #Entering Cl -[g mol]\n",
"out_Cl = pml_HCL ; # Exiting Cl in HCl-[g mol]\n",
"ef_w = out_Cl/in_Cl ; # Mole efficiency of waste\n",
"ef_pr = 1-ef_w ; # Mole efficiency of product\n",
"print ' (h) Mole efficiency of product is %.3f '%ef_pr"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Total chlorine fed to reactor is 170.10 g mol \n",
" Total C3H6 fed to reactor is 680.10 g mol \n",
" (b) Cl2 is limiting reactant \n",
" (c) C3H6 is excess reactant \n",
" (d) Fractional conversion of C3H6 to C3H5Cl is 6.76e-03 \n",
" (e) Selectivity of C3H5Cl relative to C3H6Cl2 is 0.19 g mol C3H5Cl/g mol C3H6Cl2 \n",
" (f) Yield of C3H5Cl per g C3H6 fed to reactor is 0.012 g C3H5Cl/g C3H6 \n",
" (g) Extent of reaction a as C3H5Cl is produced only in reaction a is 4.6 \n",
" Extent of reaction b as C3H6Cl2 is produced only in reaction b 24.5 \n",
" (h) Mole efficiency of product is 0.986 \n"
]
}
],
"prompt_number": 8
},
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"cell_type": "code",
"collapsed": true,
"input": [],
"language": "python",
"metadata": {},
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}
],
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}
]
}