When gases B, C and D were passed through a tube of powdered , 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 at

Question

When gases B, C and D were passed through a tube of powdered $Li_2O$, 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?

Prashant Kotian · Accepted Answer

Correct answer: B. At 300 K (27°C), the moles of remaining gases C and D are: n_CD = PV / RT = 2.1 × 0.821 / (0.0821 × 300) = 1.7241 / 24.63 = 0.07 mol. Since the moles of B, C, D combined was 0.19 mol, the moles of B absorbed is 0.19 - 0.07 = 0.12 mol. This gas B is , which reacts with to form Li2CO3.

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