Answer to Question #193635 in Chemistry for Valentina

Question

How many grams of chlorine gas, Cl2, are in 3.12 L at -14 ° C and 253 kPa pressure?

Answer 1

P = 253 kPa

V = 3.12 L

R = 8.314  kPa L K^-1 mol^-1

T = -14°C = 259 K

Solution:

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.

Therefore,

(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).