Equal masses of Na2O2 and KO2 are separately treated with excess CO2 (giving the metal carbonate and O2). What is the ratio of volume of O2 from KO2 to that from Na2O2 under identical conditions? (Na = 23, K = 39, O = 16)

JEE Advanced levelSingle correct optionDifficulty 3 of 3sb004

Equal masses of $Na X_{2} O X_{2}$ and $KO X_{2}$ are separately treated with excess $CO X_{2}$ (giving the metal carbonate and $O X_{2}$ ). What is the ratio of volume of $O X_{2}$ from $KO X_{2}$ to that from $Na X_{2} O X_{2}$ under identical conditions? (Na = 23, K = 39, O = 16)

2 Na X_{2} O X_{2} + 2 CO X_{2} 2 Na X_{2} CO X_{3} + O X_{2}

gives

0.5

mol

O X_{2}

per mole of peroxide (molar mass 78).

4 KO X_{2} + 2 CO X_{2} 2 K X_{2} CO X_{3} + 3 O X_{2}

gives

0.75

mol

O X_{2}

per mole of

KO X_{2}

(molar mass 71). For

1 g

0.75/71 = 0.01056

and

0.5/78 = 0.00641

mol, so the ratio is

1.65