Potassium chlorate (KClO3, 122.5 g/mol) is heated with MnO2 as catalyst: 2KClO3 -> 2KCl + 3O2. The volume of oxygen at STP (22.4 L/mol) from 12.25 g of KClO3 is

JEE Main levelSingle correct optionDifficulty 2 of 3pn187

Potassium chlorate ( $KClO X_{3}$ , 122.5 g/mol) is heated with $MnO X_{2}$ as catalyst: $2 KClO X_{3} 2 KCl + 3 O X_{2}$ . The volume of oxygen at STP (22.4 L/mol) from 12.25 g of $KClO X_{3}$ is

Moles of

KClO X_{3}

= 12.25/122.5 = 0.10. Each mole gives 1.5 mol

O X_{2}

, so 0.15 mol forms. Volume =

0.15 \times 22.4 = 3.36