The volume of CO₂ obtained at STP by the complete decomposition of 9.85 g Na₂CO₃ is:

(A) 2.25 litre

(B) zero

(C) 0.85 litre

(D) 0.56 litre

Given:

• Mass of Na₂CO₃ = 9.85 g
• Molar mass of Na₂CO₃ = 106 g/mol

Step 1: Calculate the number of moles of Na₂CO₃

The number of moles is calculated using the formula:

Number of moles = Mass of substance / Molar mass

Substitute the values:

Number of moles of Na₂CO₃ = 9.85 g / 106 g/mol = 0.0929 mol

Step 2: Decomposition Reaction of Na₂CO₃

The decomposition reaction of Na₂CO₃ is:

Na₂CO₃ → Na₂O + CO₂

From the equation, 1 mole of Na₂CO₃ produces 1 mole of CO₂. So, the number of moles of CO₂ is:

Number of moles of CO₂ = 0.0929 mol

Step 3: Calculate the Volume of CO₂ at STP

At STP, 1 mole of gas occupies 22.4 liters. So, the volume of CO₂ produced is:

Volume of CO₂ = Number of moles of CO₂ × 22.4 L/mol

Substituting the values:

Volume of CO₂ = 0.0929 mol × 22.4 L/mol = 2.08 liters

Final Answer

Therefore, the volume of CO₂ obtained is approximately 2.08 liters. The closest answer in the options is:

Correct answer: (A) 2.25 liters