what is the volume occupied by 16.0g ethane gas (C2H6) at 720 Torr and 18C?

Question

V=nRT/P

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

Answer 1

I don't get that number exactly. First, I don't think you should have rounded when you did. You are allowed at least 3 s.f. but you rounded to 0.53, throwing away at least 1 s.f. I obtained, using your numbers, 13.37 L, which would be rounded to 13.4 to three s.f. However, using the number in the calculator that comes from 16/30.07 (0.53209), and using the remainder of the figures you have, returns 13.418 which rounds to 13.4L.

Answer 2

I guess I need to figure out the s.f. or stop rounding numbers.... :( thank you...

Answer 3

I ALWAYS carry at least one more place than I know I'm allowed and sometimes even two more. You can ALWAYS throw those unwanted numbers away at the end when you round with s.f. in mind; you can't always ADD numbers at the end if you've thrown them away earlier.