Answer
$4C_2H_5NH_2(g)+15O_2(g)\rightarrow 8CO_2(g)+14H_2O(g)+2N_2(g)$
Work Step by Step
$C_2H_5NH_2(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)+N_2(g)$
Add a coefficient of 2 to $C_2H_5NH_2$ to create an even number of $H$ atoms on the left. Since $H$ only appears in pairs on the right, the number we are balancing to must be even.
$2C_2H_5NH_2(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)+N_2(g)$
Add a coefficient of 4 to $CO_2$ to balance $C$.
$2C_2H_5NH_2(g)+O_2(g)\rightarrow 4CO_2(g)+H_2O(g)+N_2(g)$
There are now 14 $H$ on the left and only 2 on the right. Add a coefficient of 7 to $H_2O$ to balance $H$.
$2C_2H_5NH_2(g)+O_2(g)\rightarrow 4CO_2(g)+7H_2O(g)+N_2(g)$
There are now an odd number of $O$ atoms on the right. $O$ only appears in pairs on the left side. Double all coefficients to elimnate the negative.
$4C_2H_5NH_2(g)+2O_2(g)\rightarrow 8CO_2(g)+14H_2O(g)+2N_2(g)$
There are now 30 $O$ atoms on the left and only 4 on the left. Increase the coefficient of $O_2$ from 2 to 15 to provide the 16 additional $O$ atoms.
$4C_2H_5NH_2(g)+15O_2(g)\rightarrow 8CO_2(g)+14H_2O(g)+2N_2(g)$
