When gases B, C and D were passed through a tube of powdered Li₂O, gas B reacted to form Li₂CO₃. The — States of Matter and Gaseous State Chemistry Question
Question
When gases B, C and D were passed through a tube of powdered Li₂O, gas B reacted to form Li₂CO₃. The remaining gases, C and D, were collected in another 821 ml flask and found to have a pressure of 2.1 atm at 27°C. How many moles of B were present and what is its likely identity?
Answer: B
💡 Solution & Explanation
B forms carbonate, so it is CO₂. Remaining C+D at 300 K: $n = (2.1 \times 0.821) / (0.0821 \times 300) = 0.07 \text{ mol}$. Moles of B = $0.19 \text{ (from prev)} - 0.07 = 0.12 \text{ mol}$. Identity is CO₂. Therefore, correct answer is B.
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