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:
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- Statement I is correct but Statement II is incorrect.
- Both Statement I and Statement II are incorrect.
- Statement I is incorrect but Statement II is correct.
- Both Statement I and Statement II are correct.
The Correct Option is A
Solution and Explanation
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:
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:
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.
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