Answer
$\dot{V}=10.38\ L/min$
Work Step by Step
Using stoichiometry, we find:
$CH_4+2\ O_2\rightarrow CO_2+2\ H_2O$
Molar flowrate of methane:
$\dot{n}=P\dot{V}/RT$
$\dot{n}=(773/760×1\ atm)×5.0\ L/min/((0.082)((28+273)))$
$\dot{n}=0.207\ mol/min$
From stoichiometry, the rate for oxygen is:
$\dot{n}=0.413\ mol/min$
Oxygen volume flowrate:
$\dot{V}=\dot{n}RT/P$
$\dot{V}=0.413\ mol/min×0.082×(26+273)\ K/(742/760×1\ atm)$
$\dot{V}=10.38\ L/min$
