Question #877f4

1 Answer
Kris Caceres
Dec 1, 2017

2.76 g of #CH_4#

Explanation:

V = 3.86 L; M = 16.0 #g/(mol)#

@ STP, P = 1.00 atm; T = 273.15 K

#PV =nRT#

#g CH_4 = (PV) /(RT) * M#

#g CH_4 = (1.00 atm * 3.86 L) /(0.08206 (L-atm)/(mol*K) * 273.15 K) * 16 g/(mol)#

#g CH_4 = 2.76 g#

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