A mixture of 0.20 mol KClO3 and 0.20 mol KMnO4 is heated strongly until both decompose completely (2KClO3 -> 2KCl + 3O2 and 2KMnO4 -> K2MnO4 + MnO2 + O2). The volume of O2 at STP (22.4 L/mol) is

JEE Advanced levelSingle correct optionDifficulty 3 of 3pn200

A mixture of 0.20 mol $KClO X_{3}$ and 0.20 mol $KMnO X_{4}$ is heated strongly until both decompose completely ( $2 KClO X_{3} 2 KCl + 3 O X_{2}$ and $2 KMnO X_{4} K X_{2} MnO X_{4} + MnO X_{2} + O X_{2}$ ). The volume of $O X_{2}$ at STP (22.4 L/mol) is

KClO X_{3}

gives

0.20 \times 1.5 = 0.30

mol

O X_{2}

KMnO X_{4}

gives

0.20/2 = 0.10

mol. Total 0.40 mol, volume

= 0.40 \times 22.4 = 8.96