conversion of mol:
16.0g C2H6 X 1mol C2H6/30.07g C2H6= .53mol C2H6
V= .53 mol(0.0821 L*atm/mol*K)(291)K /
720 Torr (1 atm/760Torr)
V= 13.5 L
Here is a quick check. You have a half mole at near stp. About 11 liters. But in reality, your pressure is about ten percent lower than stp, so volume about 12 liters, and temp is about 5 percent higher, so volume is about 13 liters.
I would go with your answer.
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