Problem: 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 CO2(g) and H2O(g). Calculate the volume of liquid H2O formed if the reaction goes to completion and you condense the water vapor.

FREE Expert Solution

CH₃OH (l) + 2 O₂ (g) → 2 H₂O (l) + CO₂ (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 H₂O

from oxygen gas

$$\begin{gathered} \mathbf{n}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{PV}}{\mathbf{RT}} \\ \mathbf{n}\mathbf{ }\mathbf{=}\mathbf{ }\frac{(1 \mathrm{atm})(13.0 L)}{(0.08206 {\displaystyle \frac{\mathrm{atm}·L}{\mathrm{mol}·K}})(273.15 K)} \\ \mathbf{n}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{58}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}\mathbf{\times }\frac{\mathbf{2}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}{\mathbf{2}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}} \end{gathered}$$

n = 0.58 mol H₂O