How many liters of space will 7.80 moles of methane gas (CH4) occupy at STP

V CH4 (g) = 190.6 L

Explanation:

assuming ideal gas:

PV = RTn

∴ STP: T = 298 K and P = 1 atm

∴ R = 0.082 atm. L/K. mol

∴ moles (n) = 7.80 mol CH4 (g)

∴ Volume CH4 (g) = ?

⇒ V = RTn/P

⇒ V CH4 (g) = ((0.082 atm. L/K. mol) * (298 K) * (7.80 mol)) / (1 atm)

⇒ V CH4 (g) = 190.6 L