Question

What will be the volume of oxygen gas produced, If the reaction
2 KClO\(_3\)(s) \(\rightarrow\) 2 KCl(s) + 3 O\(_2\)(g) \(\Delta H^\circ = -78 \, \text{kJ}\)
is carried out at S.T.P.?

• A. 48.0 L
• B. 44.8 L
• C. 22.4 L
• D. 67.2 L

Hint:

Remember that at STP, 1 mole of gas occupies 22.4 L, and use stoichiometry to calculate volumes of gases in reactions.

Answer 1

The Correct Option is D

Step 1: Understanding the reaction.
The balanced reaction shows that 2 moles of KClO\(_3\) decompose to give 3 moles of O\(_2\). We need to calculate the volume of oxygen gas produced at standard temperature and pressure (STP). At STP, 1 mole of any ideal gas occupies 22.4 L.

Step 2: Calculating the volume.
The volume of oxygen produced is directly proportional to the amount of KClO\(_3\) used. The given reaction shows that 2 moles of KClO\(_3\) produce 3 moles of O\(_2\). Therefore, for every 2 moles of KClO\(_3\), 3 moles of O\(_2\) are produced, which corresponds to: \[ \text{Volume of O}_2 = \frac{3}{2} \times 22.4 \, \text{L} = 67.2 \, \text{L}. \]
Step 3: Conclusion.
Thus, the correct answer is (D) 67.2 L, as this is the volume of oxygen gas produced at STP.