Question

The right option for the mass of CO₂ produced by heating 20g of 20% pure limestone is 
(Atomic mass of Ca= 40) 

• A. 1.76 g
• B. 2.64 g
• C. 1.32 g
• D. 1.12 g

Answer 1

The Correct Option is A

To determine the mass of CO₂ produced by heating 20g of 20% pure limestone (CaCO₃), we follow these steps:

1. First, calculate the mass of CaCO₃ present in the limestone.
   Since the limestone is 20% pure:
   Mass of CaCO₃ = 20% of 20g = (20/100) × 20 = 4g
2. Next, calculate the moles of CaCO₃.
   The molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 16 × 3 (O) = 100 g/mol.
   Moles of CaCO₃ = Mass / Molar mass = 4g / 100 g/mol = 0.04 mol
3. From the reaction: CaCO₃ → CaO + CO₂,
   1 mol of CaCO₃ produces 1 mol of CO₂. Therefore, 0.04 mol of CaCO₃ will produce 0.04 mol of CO₂.
4. Calculate the mass of CO₂ produced.
   The molar mass of CO₂ = 12 (C) + 16 × 2 (O) = 44 g/mol.
   Mass of CO₂ = Moles × Molar mass = 0.04 mol × 44 g/mol = 1.76 g

Thus, the mass of CO₂ produced is 1.76 g.

Answer 2

Molecular mass of CaCO₃ = 40+12+3x16 = 100 g/mol
The limestone is 20% pure, which means only 20% of it is calcium carbonate.

Moles of CaCO₃ = \(\frac {Mass\  of\  CaCO_3}{Molecular\  mass \ of\  CaCO_3 }\)= \(\frac {20 g \times 0.20}{100 \ g/mol}\) = 0.04 moles

Determine the moles of CO₂ produced. From the balanced chemical equation, 1 mole of CaCO₃ produces 1 mole of CO₂. The molar ratio of CaCO₃ to CO₂ is 1:1.
Therefore, moles of CO₂ produced = moles of CaCO₃ = 0.04 moles
Molecular mass of CO₂ = 12+2x16 = 44 g/mol

Mass of CO₂ produced = (moles of CO₂) x (molecular mass of CO₂) = 0.04 moles x 44 g/mol = 1.76 g
So, the correct option is (A): 1.76 g