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+// Example 12_4
+clc;funcprot(0);
+// Given data
+// The volumetric analysis of the products on dry basis
+CO_2=10.4// %
+CO=1.2;// %
+O_2=2.8;// %
+N_2=85.6;// %
+
+// Calculation
+// The chemical equation is C_aH_b+c(O_2+3.76N_2)-->10.4CO_2+1.2CO+2.8O_2+85.6N_2+dH_2O
+// Balancing each element,
+a=10.4+1.2;// (C)
+c=85.6/3.76;// (N)
+d=(2*c)-(20.8+1.2+5.6);// (O)
+b=2*d;// (H)
+printf("\nThe chemical formula for the fuel is C_%2.1fH_%2.1f",a,b);
+// The find the percent theoretical air from the actual chemical equation, C_11.6H_37.9+21.08(O_2+3.76N_2)-->11.6CO_2+18.95H_2O+79.26N_2
+c_act=21.08;
+P_ta=(c/c_act)*100;// The percent theoretical air in %
+printf("\nThe percent theoretical air=%3.1f percentage",P_ta);