Statement I

“One mole of propyne reacts with excess of sodium to liberate half a mole of $\mathrm{H}_2$ gas.”

Reaction and Stoichiometry

Propyne (terminal alkyne): $\mathrm{CH_3C \equiv CH}$.

Reaction with sodium:

$ 2\,\mathrm{CH_3C \equiv CH} \;+\; 2\,\mathrm{Na} \;\longrightarrow\; 2\,(\mathrm{CH_3C \equiv C^-Na^+}) \;+\; \mathrm{H_2}. $

From this balanced equation, 2 moles of propyne produce 1 mole of $\mathrm{H_2}$.

Hence, 1 mole of propyne will produce $\tfrac{1}{2}$ mole of $\mathrm{H_2}$.

$ \boxed{\text{Statement I is correct.}} $

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Statement II

“Four grams of propyne reacts with $\mathrm{NaNH_2}$ to liberate $\mathrm{NH_3}$ gas which occupies 224 mL at STP.”

Analysis

Moles of propyne

Molecular mass of propyne ($\mathrm{C_3H_4}$):

$ 3 \times 12 + 4 \times 1 = 36 + 4 = 40\,\mathrm{g/mol}. $

Four grams of propyne is:

$ \frac{4\,\mathrm{g}}{40\,\mathrm{g/mol}} = 0.1\,\mathrm{mol}. $

Reaction with $\mathrm{NaNH_2}$

For a terminal alkyne:

$ \mathrm{CH_3C \equiv CH} \;+\; \mathrm{NaNH_2} \;\longrightarrow\; \mathrm{CH_3C \equiv C^-Na^+} \;+\; \mathrm{NH_3}. $

1 mole of propyne produces 1 mole of $\mathrm{NH_3}$.

Moles of $\mathrm{NH_3}$ produced

With $0.1\,\mathrm{mol}$ of propyne, we get $0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$.

Volume of $\mathrm{NH_3}$ at STP

1 mole of any ideal gas at STP $\approx 22.4\,\mathrm{L} = 22400\,\mathrm{mL}.$

$0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$ occupies $0.1 \times 22.4\,\mathrm{L} = 2.24\,\mathrm{L} = 2240\,\mathrm{mL}.$

However, Statement II says the liberated $\mathrm{NH_3}$ occupies only 224 mL at STP, which corresponds to $0.01\,\mathrm{mol}$ of $\mathrm{NH_3}$, not $0.1\,\mathrm{mol}$. Therefore, the statement’s volume is off by a factor of 10 and is thus incorrect if the reaction goes to completion in a typical way.

$ \boxed{\text{Statement II is incorrect.}} $

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Conclusion

Statement I is correct.

Statement II is incorrect.

Hence, the best choice is:

$ \boxed{\text{Option D: Statement I is correct but Statement II is incorrect.}} $