Given below are two statements :

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

Statement II : Four g of propyne reacts with $\mathrm{NaNH}_2$ to liberate $\mathrm{NH}_3$ gas which occupies 224 mL at STP.

In the light of the above statements, choose the most appropriate answer from the options given below :

<h2><strong>Statement I</strong></h2> <p>“One mole of propyne reacts with excess of sodium to liberate half a mole of $\mathrm{H}_2$ gas.”</p> <h3><strong>Reaction and Stoichiometry</strong></h3> <p><p>Propyne (terminal alkyne): $\mathrm{CH_3C \equiv CH}$. </p></p> <p><p>Reaction with sodium: </p> <p>$ 2\,\mathrm{CH_3C \equiv CH} \;+\; 2\,\mathrm{Na} \;\longrightarrow\; 2\,(\mathrm{CH_3C \equiv C^-Na^+}) \;+\; \mathrm{H_2}. $ </p></p> <p><p>From this balanced equation, <strong>2 moles</strong> of propyne produce <strong>1 mole</strong> of $\mathrm{H_2}$. </p></p> <p>Hence, <strong>1 mole</strong> of propyne will produce <strong>$\tfrac{1}{2}$ mole</strong> of $\mathrm{H_2}$. </p> <p>$ \boxed{\text{Statement I is correct.}} $</p> <hr /> <h2><strong>Statement II</strong></h2> <p>“Four grams of propyne reacts with $\mathrm{NaNH_2}$ to liberate $\mathrm{NH_3}$ gas which occupies 224 mL at STP.”</p> <h3><strong>Analysis</strong></h3> <p><strong>Moles of propyne</strong> </p> <p><p>Molecular mass of propyne ($\mathrm{C_3H_4}$): </p> <p>$ 3 \times 12 + 4 \times 1 = 36 + 4 = 40\,\mathrm{g/mol}. $</p></p> <p><p>Four grams of propyne is: </p> <p>$ \frac{4\,\mathrm{g}}{40\,\mathrm{g/mol}} = 0.1\,\mathrm{mol}. $</p></p> <p><strong>Reaction with $\mathrm{NaNH_2}$</strong> </p> <p><p>For a <strong>terminal alkyne</strong>: </p> <p>$ \mathrm{CH_3C \equiv CH} \;+\; \mathrm{NaNH_2} \;\longrightarrow\; \mathrm{CH_3C \equiv C^-Na^+} \;+\; \mathrm{NH_3}. $</p></p> <p><p><strong>1 mole</strong> of propyne produces <strong>1 mole</strong> of $\mathrm{NH_3}$.</p></p> <p><strong>Moles of $\mathrm{NH_3}$ produced</strong> </p> <p>With $0.1\,\mathrm{mol}$ of propyne, we get $0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$.</p> <p><strong>Volume of $\mathrm{NH_3}$ at STP</strong> </p> <p><p>1 mole of any ideal gas at STP $\approx 22.4\,\mathrm{L} = 22400\,\mathrm{mL}.$ </p></p> <p><p>$0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$ occupies $0.1 \times 22.4\,\mathrm{L} = 2.24\,\mathrm{L} = 2240\,\mathrm{mL}.$</p></p> <p>However, <strong>Statement II</strong> says the liberated $\mathrm{NH_3}$ occupies only <strong>224 mL</strong> 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 <strong>incorrect</strong> if the reaction goes to completion in a typical way.</p> <p>$ \boxed{\text{Statement II is incorrect.}} $</p> <hr /> <h2><strong>Conclusion</strong></h2> <p><p><strong>Statement I</strong> is <strong>correct</strong>. </p></p> <p><p><strong>Statement II</strong> is <strong>incorrect</strong>. </p></p> <p>Hence, the best choice is:</p> <p>$ \boxed{\text{Option D: Statement I is correct but Statement II is incorrect.}} $</p>