$13\mathrm{.}6.$ In Example $13\mathrm{.}3$ we assumed that all the HC was combusted, and thus all the oxygen deficit

appeared as CO$.$ The calculated CO concentration was $(2(\%)/(1\mathrm{.}1)\%)=1\mathrm{.}8$ times ~~$2$ times

what one observes in Fig. $13\mathrm{.}2.$ This suggests that the oxygen deficit is shared roughly

equally by unconverted CO and unburned HC$.$ To test this idea:

$($a$)$ Read the HC concentration for $\backslash$lambda $=0\mathrm{.}95$ from Fig. $13\mathrm{.}2.$

$($b$)$ Write the equation equivalent to Eq$.(13\mathrm{.}6)$ taking into account the possibility that

some HC is not combusted. Here show on the right a term $\backslash$alpha C$\_($x$)$H$\_($y$),$ where $\backslash$alpha represents

the mols of unburned HC$.$

$($c$)$ Then write the oxygen balance equivalent to Eq$.(13\mathrm{.}7),$ finding that it is one equation

with two unknowns, z and $\backslash$alpha $.$

$($d$)$ Solve it for $\backslash$alpha $,$ on the assumptions that z is about half the value calculated in Example

$13\mathrm{.}3$ and that the unburned HC has the same composition as the fuel, C$\_(8)$H$\_(17).$

$($e$)$ Compare that result to the one found in part $($a$).$