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+clear ;
+clc;
+// Example 9.5
+printf('Example 9.5\n\n');
+// Page no. 238
+// Solution
+
+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]
+
+// 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
+printf(' (a) Max. amount of NH3 that can be produced is %.1f g\n',mx_NH3);
+
+//(b) and (c)
+if (ex_H2 > ex_N2 )
+ printf('  (b) N2 is limiting reactant  \n');
+ printf('  (c) H2 is excess reactant  \n');
+ ex_r = ex_N2;
+  else
+printf('  (b) H2 is limiting reactant  \n');
+ printf('  (c) N2 is excess reactant  \n');
+ ex_r = ex_H2 ;
+ end
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