Answer: 10.06 g
Step-by-step explanation:
PV = nRT,
substitute the values into the equation
At STP, pressure (P) is 1 atm and temperature (T) is 273 K. The value of R is 0.0821 L*atm/mol*K.
PV = nRT
(1 atm) (15.9 L) = n (0.0821 L•atm/mol•K) (273 K)
solve for n
n = (1 atm x 15.9 L) / (0.0821 L•atm/mol•K x 273 K)
n = 0.627 moles of methane
caluclate using its molar mass
mass = n * molar mass
mass = 0.627 mol x 16.04 g/mol
mass = 10.06 g