Question

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:

• A. Statement I is correct but Statement II is incorrect.
• B. Both Statement I and Statement II are incorrect.
• C. Statement I is incorrect but Statement II is correct.
• D. Both Statement I and Statement II are correct.

Hint:

Remember to balance the chemical equations and use the molar mass to calculate the number of moles. Also, one mole of any gas occupies 22.4 L at STP.

Answer 1

The Correct Option is A

To determine the accuracy of Statements I and II, let's analyze each one:

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

The reaction of propyne (C₃H₄) with sodium (Na) involves the acidic hydrogen present in the terminal alkyne. The reaction is:

C₃H₃ - H + Na → C₃H₃Na + 1/2 H₂

In this reaction, one mole of propyne reacts with sodium to form sodium propyne and release half a mole of hydrogen gas. Therefore, Statement I is correct.

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

First, calculate the amount of propyne:

$\frac{4}{40}=0.1mol$

When propyne reacts with NaNH₂, the hydrogen atom is replaced, and NH₃ gas is formed. According to stoichiometry, 1 mole of NaNH₂ liberates 1 mole of NH₃:

C₃H₄ + NaNH₂ → C₃H₃Na + NH₃

The molar volume of gas at STP is 22.4 L/mol. Hence, the volume occupied by 0.1 mol NH₃ at STP is:

$0.1mosubmaltiplier="x">22.4=2.24L$

The calculated volume of NH₃ is 2.24 L, not 224 mL. Hence, Statement II is incorrect.

In conclusion, the correct choice is: Statement I is correct but Statement II is incorrect.