Answer to Question #193635 in Chemistry for Valentina

Question #193635

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


1
Expert's answer
2021-05-17T04:08:27-0400

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).

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