Answer :
The moles of chlorine gas in a 10.0 L tank at 27°C and 3.50 atm is approximately 1.42 moles, which corresponds to about 101 grams of Cl2.
To calculate the moles of chlorine gas (Cl2) in a 10.0 L tank at 27°C and 3.50 atm, we use the ideal gas law PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L*atm/K*mol), and T is the temperature in Kelvin.
First, we convert the temperature to Kelvin (K):
T(27°C) = 27 + 273.15
= 300.15 K.
Now, we can solve for n:
n = (PV) / (RT)
= (3.50 atm * 10.0 L) / (0.0821 L*atm/K*mol * 300.15 K)
n ≈ 1.42 moles
Using the molar mass of chlorine (70.90 g/mol), we convert moles to grams:
Mass = moles * molar mass
= 1.42 moles * 70.90 g/mol
= 100.68 grams, which is closest to the answer choice (d) 101 grams.
