Question #877f4

1 Answer
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