50g of a good sample of CaOCl₂ is made to react with CO₂.  The volume of Cl₂ liberated at S.T.P. is ____ lit.

To determine the volume of Cl₂ gas released when 50g of a sample of CaOCl₂ reacts with CO₂, let's analyze the reaction and calculations step by step.

Chemical Reaction

The reaction between calcium oxychloride (CaOCl₂) and carbon dioxide (CO₂) is as follows:

CaOCl₂ + CO₂ → CaCO₃ + Cl₂

Stoichiometric Analysis

From the balanced chemical equation, it is clear that:

• 1 mole of CaOCl₂ produces 1 mole of Cl₂.

Calculating Moles of CaOCl₂

The number of moles of CaOCl₂ in the given sample is calculated using the formula:

Moles = Mass / Molar Mass

Substituting the values:

Moles of CaOCl₂ = 50 / 198 = 0.25 moles

Determining Moles of Cl₂ Produced

Since 1 mole of CaOCl₂ liberates 1 mole of Cl₂,

Moles of Cl₂ = 0.25 moles

Calculating Volume of Cl₂ at STP

At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore:

• Volume of 1 mole of Cl₂ = 22.4 liters
• Volume of 0.25 moles of Cl₂ = 22.4 × 0.25 = 5.6 liters

Final Answer

The volume of Cl₂ gas liberated at STP when 50g of CaOCl₂ reacts with CO₂ is 5.6 liters.