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+// Example 12_2
+clc;funcprot(0);
+// Given data
+P_ta=90;// % theoretical air
+// The reaction equation for theoretical air is C_4H_10+(0.9)(6.5)(O_2+3.76N_2)-->4CO_2+5H_2O+22N_2+bCO
+a_1=6.5;// The stoichiometric coefficient
+M_air=29;// kg/kmol
+M_fuel=58;// kg/kmol
+
+// Calculation
+function[X]=atomicbalances(y)
+    X(1)=y(1)+y(2)-4;
+    X(2)=(2*y(1))+5+y(2)-11.7;
+endfunction
+y=[1 1];
+z=fsolve(y,atomicbalances);
+a=z(1);// mol
+b=z(2);// mol
+P_CO=(b/31)*100;// % CO
+m_air=(P_ta/100)*a_1*(4.76)*M_air;// lbm air
+m_fuel=1*M_fuel;// lbm fuel
+AF=m_air/m_fuel;// The air-fuel ratio in lbm air/lbm fuel
+printf("\nThe volume percentage of CO=%1.2f percentage \nThe air-fuel ratio,AF=%2.2f lbm air/lbm fuel",P_CO,AF);