What volume of oxygen gas at NTP is necessary for complete combustion of 20 litre of proportional measured at `27^(@)C` and 760mm preesure?

Explanation

The balanced equation is:
`underset(1"litre")underset(1"vol")(C_(3)H_(8))+underset(5"litre")underset(5"vol.")(5O_(2)) to 3CO_(2)(g)+4H_(2)O`
1 litre of propane requires=5litre of oxygen
20litre of propane will requires=`5xx20=100"litre"` of oxygen at 760mm pressure and `27^(@)C`
This volume of propane will be converted to NTP conditions.
`{:(,"Given conditions","NTP conditions"),(,P_(1)=760mm,P_(2)=760mm),(,V_(1)=100"litre",V_(2)=?),(,T_(1)=27+273=360K,T_(2)=273):}`
`"Applying"(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`V_(2)=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2))`
`=(760xx100)/(300)xx(273)/(760)=91.0"litre"`