2 CH_{4} + O_{2} → 2 CH_{3}OH

**$\mathbf{PV}\mathbf{}\mathbf{=}\mathbf{}\mathbf{nRT}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{PV}}{\mathbf{RT}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{atm}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{07}\mathbf{\times}{\mathbf{10}}^{\mathbf{10}}\mathbf{}{\mathbf{ft}}^{\mathbf{3}}\mathbf{}\mathbf{\times}{\displaystyle \frac{28.317L}{1{\mathrm{ft}}^{3}}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}\mathbf{}{\displaystyle \frac{\mathrm{atm}\xb7L}{\mathrm{mol}\xb7K}}\mathbf{)}\mathbf{(}\mathbf{25}\mathbf{\xb0}\mathbf{C}\mathbf{}\mathbf{+}\mathbf{}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{)}\mathbf{K}}$**

**n = 1.2384x10 ^{10} mol**

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 ^{o}C. One possible strategy is to oxidize the methane to methanol, CH_{3}OH, which has a boiling point of 65 ^{o}C and can therefore be shipped more readily. Suppose that 1.07×10^{10} ft^{3} of methane at atmospheric pressure and 25 ^{o}C are oxidized to methanol.

What volume of methanol is formed if the density of CH_{3}OH is 0.791 g/mL?

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