Question

Given below are two statements: Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H₂ gas. Statement II: Four g of propyne reacts with NaNH₂ to liberate NH₃ gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:

• A. Statement I is correct but Statement II is incorrect.
• B. Both Statement I and Statement II are incorrect.
• C. Statement I is incorrect but Statement II is correct.
• D. Both Statement I and Statement II are correct.

Hint:

Remember to balance the chemical equations and use the molar mass to calculate the number of moles. Also, one mole of any gas occupies 22.4 L at STP.

Answer 1

The Correct Option is A

Statement I: The reaction of propyne (CH₃-C≡CH) with sodium (Na) is: CH₃-C≡CH + Na → CH₃-C≡C⁻Na⁺ + ½H₂↑ 1 mole (excess) ½ mole H₂ One mole of propyne reacts with excess sodium to produce ½ mole of H₂ gas. Therefore, Statement I is correct. 

Statement II: The reaction of propyne (CH₃-C≡CH) with sodium amide (NaNH₂) is: CH₃-C≡CH + NaNH₂ → CH₃-C≡C⁻Na⁺ + NH₃ Molar mass of propyne (C₃H₄) = 3(12) + 4(1) = 36 + 4 = 40 g/mol 4 g of propyne = 4/40 = 0.1 mole One mole of propyne produces one mole of NH₃. 0.1 mole of propyne produces 0.1 mole of NH₃. Volume of 0.1 mole of NH₃ at STP = 0.1 × 22.4 L = 2.24 L = 2240 mL The given volume is 224 mL, which is incorrect. 

Therefore, Statement II is incorrect.