Question

Volume of \(CO_2\) obtained by the complete decomposition of \(9.85\ gm \ BaCO_3\) is:

• A. \(2.24 \ lit.\)
• B. \(1.12\ lit.\)
• C. \(0.84\ lit.\)
• D. \(0.56\ lit.\)

Answer 1

The Correct Option is B

Molar mass of Ba = 137.33 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of O = 16.00 g/mol

Molar mass of BaCO₃ = 137.33 + 12.01 + 3 x 16.00 = 197.34 g/mol

Moles of BaCO₃ = \(\frac {Mass }{Molar \ mass }\)

Moles of BaCO₃ = \(\frac {9.85\  g }{ 197.34\  g/mol}\) ≈ 0.05 moles

Balanced chemical equation for the decomposition of BaCO₃

\(BaCO_3 → BaO + CO_2\)

The stoichiometry tells us that one mole of BaCO₃ produces one mole of CO_2.

Moles of CO₂ = Moles of BaCO₃ ≈ 0.05 moles

Now, use the ideal gas law to find the volume of CO₂ at standard temperature and pressure (STP):

The molar volume of an ideal gas at STP is 22.4 liters/mol.

Volume of CO₂ = Moles of CO₂ x Molar volume at STP 

Volume of CO₂ = 0.05 moles x 22.4 liters/mole = 1.12 liters

So, the correct option is (B): \(1.12\ lit.\)