Answer:
the mass required to inflate a 72 L bag is 191.491 g
Step-by-step explanation:
reaction:
conditions:
- V = 72.0 L
- STP: P = 1 atm ∧ T = 298 K
gas law:
- PV = RTn
- R = 0.082 atm * L / K * mol
⇒ n = PV / RT
⇒ n = ((1 atm) * ( 72.0 L)) / (0.082 atm*L / K*mol) * (298 K)
⇒ n = 2.946 mol
⇒ m = n * Mw = ( 2.946 mol ) * ( 64.99 g/mol)
⇒ m = 191.491 g