JEE Main 2025 Chemistry question
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 :
Answer: (1)
Solution
1 mole propyne + Na → 1/2 mole H2 (Statement I correct). 4g propyne = 0.1 mol,
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