Question

How many grams of chlorine gas (Cl2) are in a 17.8 liter sample at 1.1 atmospheres and 29°C? Show all work used to solve this problem.

Answer 1

T = 29 + 273 = 302 K
moles Cl2 = pV/RT = 1.1 x 17.8 / 0.08206 x 302 =0.790
mass Cl2 = 0.790 mol x 70.906 g/mol=56.0 g

Answer 2

PV=nRT
n= 1.1*17.8/[0.08206*(29+273.15)]
n= 0.78969 (do not round yet)
mass= 0.78969*(35.453*2)= 55.99g = 56.0 g (3 s.f)