How many grams of chlorine gas, Cl2, are in 3.12 L at -14 ° C and 253 kPa pressure?
P = 253 kPa
V = 3.12 L
R = 8.314 kPa L K-1 mol-1
T = -14°C = 259 K
The Ideal Gas equation can be used.
The Ideal Gas equation can be expressed as: PV = nRT
To find the moles of chlorine gas, solve the equation for n:
n(Cl2) = PV / RT
n(Cl2) = (253 kPa × 3.12 L) / (8.314 kPa L K-1 mol-1 × 259 K) = 0.3666 mol
Moles of Cl2 = 0.3666 mol
The molar mass of Cl2 is 70.906 g/mol.
(0.3666 mol Cl2) × (70.906 g Cl2 / 1 mol Cl2) = 25.994 g Cl2 = 26 g Cl2
Answer: 26 grams of chlorine gas (Cl2).