**Equilibrium Reaction: C _{3}H_{8(g)} + 6H_{2}O_{(g)} → 3CO_{2(g)} + 10H_{2(g)} **

$\mathbf{mole}\mathbf{}\mathbf{fraction}\left(\chi \right)\mathbf{}{\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{}\mathbf{=}\frac{\mathbf{moles}\mathbf{}{\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}}{\mathbf{total}\mathbf{}\mathbf{moles}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mol}}{\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mol}}\mathbf{=}$**0.2**

$\mathbf{mole}\mathbf{}\mathbf{fraction}\left(\chi \right)\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{moles}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}}{\mathbf{total}\mathbf{}\mathbf{moles}}\mathbf{=}\frac{\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mol}}{\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mol}}\mathbf{=}$**0.8**

The methane used to obtain H_{2} for NH_{3} manufacture is impure and usually contains other hydrocarbons, such as propane, C_{3}H_{8}. Imagine the reaction of propane occurring in two steps:

C_{3}H_{8}(g) + 3H_{2}O(g) ⥫⥬ 3CO(g) + 7H _{2}(g)

K _{p} = 8.175×10^{15} at 1200. K

CO(g) + H_{2}O(g) ⥫⥬ CO_{2}(g) + H_{2}(g)

K _{p} = 0.6944 at 1200. K

(c) When 1.00 volume of C_{3}H_{8} and 4.00 volumes of H_{2}O, each at 1200. K and 5.0 atm, are mixed in a container, what is the final pressure? Assume the total volume remains constant, that the reaction is essentially complete, and that the gases behave ideally.

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