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Solution


Q.

Given below are two statements:                                      [2025]

Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H2 gas.

Statement II: Four gram of propyne reacts with NaNH2 to liberate NH3 gas which occupies 224 mL.

In the light of the above statements, choose the most appropriate answer from the options given below:
1 Statement I is correct but Statement II is incorrect  
2 Both Statement I and Statement II are incorrect  
3 Statement I is incorrect but Statement II is correct  
4 Both Statement I and Statement II are correct  

Ans.

(1)

Statement I: One mole propyne gives half mole hydrogen gas as per the following stoichiometry:

CH3-CPropyne≡C-H →Na (excess) CH3-C≡C(-)Na(+)+12H2

Statement II:

CH3-CPropyne≡C-H+NaNH2⟶CH3-C≡C(-)Na(+)+NH3

Moles of propyne=Given massMolar mass=4 g40 g mol-1=0.1 mol

By stoichiometry, moles of NH3 = moles of propyne = 0.1 mol.

Volume occupied by 0.1 mol gas at STP = 0.1×22400 mL = 2240 mL