## Chemistry and Chemical Reactivity (9th Edition)

$\dot{V}=10.38\ L/min$
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: $\dot{V}=\dot{n}RT/P$ $\dot{V}=0.413\ mol/min×0.082×(26+273)\ K/(742/760×1\ atm)$ $\dot{V}=10.38\ L/min$