CH_{3}OH (l) + 2 O_{2} (g) → 2 H_{2}O (l) + CO_{2} (g)

Step 1

from methanol

**$\mathbf{26}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{mLCH}}_{\mathbf{3}}\mathbf{OH}\mathbf{}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{850}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}}{\mathbf{1}\mathbf{}\mathbf{mL}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}}{\mathbf{32}\mathbf{.}\mathbf{042}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}}$**

**= 1.38 mol H _{2}O**

from oxygen gas

**$\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{PV}}{\mathbf{RT}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{(1\mathrm{atm})(13.0L)}{(0.08206{\displaystyle \frac{\mathrm{atm}\xb7L}{\mathrm{mol}\xb7K}})(273.15K)}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{58}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{O}}_{\mathbf{2}}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{O}}_{\mathbf{2}}}$**

**n = 0.58 mol H _{2}O**

Consider the combustion reaction between 26.0 mL of liquid methanol (density = 0.850 /mL) and 13.0 L of oxygen gas measured at STP. The products of the reaction are CO_{2}(g) and H_{2}O(g). Calculate the volume of liquid H_{2}O formed if the reaction goes to completion and you condense the water vapor.

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