Problem: Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and thus has a boiling point at atmospheric pressure of -164 oC. One possible strategy is to oxidize the methane to methanol, CH3OH, which has a boiling point of 65 oC and can therefore be shipped more readily. Suppose that 1.07×1010 ft3 of methane at atmospheric pressure and 25 oC are oxidized to methanol.(A) What volume of methanol is formed if the density of CH3OH is 0.791 g/mL? (B)Calculate the total enthalpy change for complete combustion of the equivalent amount of methanol, as calculated in part A.

FREE Expert Solution

2 CH₄ + O₂ → 2 CH₃OH

$$\begin{gathered} \mathbf{PV}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{nRT} \\ \mathbf{n}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{PV}}{\mathbf{RT}} \\ \mathbf{n}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{(}\mathbf{1}\mathbf{ }\mathbf{atm}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{07}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{ }{\mathbf{ft}}^{\mathbf{3}}\mathbf{ }\mathbf{\times }{\displaystyle \frac{28.317 L }{1 {\mathrm{ft}}^3}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}\mathbf{ }{\displaystyle \frac{\mathrm{atm}·L}{\mathrm{mol}·K}}\mathbf{)}\mathbf{(}\mathbf{25}\mathbf{°}\mathbf{C}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{)}\mathbf{K}} \end{gathered}$$

n = 1.2384x10¹⁰ mol

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