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: (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.

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Imagine a Lego stick (propyne)

It has a special block (the hydrogen directly attached to the triple bond) that can be popped off quite easily when it meets certain metals or bases.

1. When it meets shiny sodium metal (Na), two Lego sticks have to pop off one special block each to give one full hydrogen gas balloon (H2). That means one stick only makes half a balloon.
2. When it meets a powerful Lego-popper called sodamide (NaNH2), every stick pops off one special block and sends out one puff of ammonia gas (NH3).

If you know how many sticks you have and how many puffs or balloons each reaction makes, you can count the volume of gas using the rule: 1 mole of any gas = 22.4 L at STP. That’s like saying every full balloon has the same size under standard conditions.

Step 1: Molar mass of propyne

Propyne $\;(CH_3 – C \equiv CH)$ has the formula $C_3H_4$.

$M = 3 \times 12 + 4 \times 1 = 36 + 4 = 40\,\text{g mol}^{-1}$

Statement I

Reaction with metallic sodium:

$2\,C_3H_4 + 2\,Na \;\longrightarrow\; 2\,C_3H_3Na + H_2$

From the equation: $2\,\text{mol}$ propyne → $1\,\text{mol}$ $H_2$.
Therefore $1\,\text{mol}$ propyne → $\frac{1}{2}\,\text{mol}$ $H_2$.
Statement I is correct.

Statement II

1. Given mass of propyne:

   $n = \frac{4\,\text{g}}{40\,\text{g mol}^{-1}} = 0.10\,\text{mol}$

2. Reaction with $NaNH_2$:

   $C_3H_4 + NaNH_2 \longrightarrow C_3H_3Na + NH_3$

   Thus 1 mol propyne → 1 mol $NH_3$.

   $0.10\,\text{mol propyne} \Rightarrow 0.10\,\text{mol } NH_3$

3. Volume of $0.10\,\text{mol}$ $NH_3$ at STP:

   $V = n \times 22.4\,\text{L} = 0.10 \times 22.4 = 2.24\,\text{L} = 2240\,\text{mL}$

   The statement claims 224 mL, which corresponds to only 0.01 mol $NH_3$. Therefore Statement II is incorrect.

Final choice

Only Statement I is correct → Option (1).