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diff --git a/1808/CH2/EX2.7/Chapter2_Example7.sce b/1808/CH2/EX2.7/Chapter2_Example7.sce
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+clc
+clear
+//INPUT DATA
+//aC4H10+ b O2+c N2= 7.8 CO2+1.1 CO +8.2 O2+82.9 N2+d H2O ;//Combustion equation
+c=82.9;//Nitrogen balance
+b=22.04;//(c/b)=3.76
+a=2.22;//Carbon balance
+d=11.1;//Hydrogen balance
+Ma=29;//mass of air
+
+//2.22C4H10+ 22.04 O2+82.9 N2= 7.8 CO2+1.1 CO +8.2 O2+82.9 N2+11.1 H2O ;//Combustion equation
+
+//CALCULATIONS
+//C4H10+ 9.92 O2+37.37 N2= 3.51 CO2+0.495 CO +3.69 O2+37.37 N2+5 H2O ;//Combustion equation dividing by 2.22 yields one mole of fuel
+xm=((9.92+37.37)*Ma)/(12*4+10);//air fuel ratio on mass basis
+
+//C4H10+ 6.5 O2+(6.5*3.76) N2= 4 CO2++24.44 N2+5 H2O ;//Theoretical Combustion equation
+xth=((6.5+24.44)*Ma)/(12*4+10);//Theoretical air fuel ratio 
+nth=(xm/xth)*100;//Percentage of theoretical air
+
+//OUTPUT
+printf('(a)Percentage of theoretical air %3.2f percentage',nth) 
+
+
+
+