Respuesta :

Answer:

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

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