Question

The reactions which produce \(\mathrm{O_2}\) are \[ \begin{aligned} \text{I. }& 2\mathrm{KClO_3}\\ \xrightarrow[\mathrm{MnO_2}]{\Delta} \\ \text{II. }& 2\mathrm{KMnO_4}\\ \xrightarrow{\Delta} \\ \text{III. }&(\mathrm{NH_4})_2\mathrm{Cr_2O_7} \xrightarrow{\Delta} \end{aligned} \] The correct answer is

• A. I, III only
• B. I, II, III
• C. II, III only
• D. I, II only

Hint:

Remember the common laboratory preparations of oxygen: \[ \boxed{ \mathrm{KClO_3} \quad\text{and}\quad \mathrm{KMnO_4} } \] on heating liberate \(\mathrm{O_2}\).

Answer 1

The Correct Option is D

Step 1: Examine Reaction I. \[ 2\mathrm{KClO_3} \xrightarrow[\mathrm{MnO_2}]{\Delta} 2\mathrm{KCl}+3\mathrm{O_2} \] Hence, \[ \boxed{\mathrm{O_2}\text{ is produced}.} \]

Step 2: Examine Reaction II. \[ 2\mathrm{KMnO_4} \xrightarrow{\Delta} \mathrm{K_2MnO_4} +\mathrm{MnO_2} +\mathrm{O_2} \] Hence, \[ \boxed{\mathrm{O_2}\text{ is produced}.} \]

Step 3: Examine Reaction III. \[ (\mathrm{NH_4})_2\mathrm{Cr_2O_7} \xrightarrow{\Delta} \mathrm{Cr_2O_3} +\mathrm{N_2} +4\mathrm{H_2O} \] No oxygen gas is evolved. Therefore, \[ \boxed{\text{Only I and II produce } \mathrm{O_2}.} \] Hence, \[ \boxed{(D)} \] is the correct answer.