A mixture of 5.85 g of Na2O2 and 6.39 g of KO2 is treated with excess CO2. Molar masses: Na2O2 78, KO2 71 g/mol; molar volume at STP is 22.4 L/mol. Calculate the volume of O2 released at STP, in litres, to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3sb154

A mixture of $5.85 g$ of $Na X_{2} O X_{2}$ and $6.39 g$ of $KO X_{2}$ is treated with excess $CO X_{2}$ . Molar masses: $Na X_{2} O X_{2}$ 78, $KO X_{2}$ 71 g/mol; molar volume at STP is 22.4 L/mol. Calculate the volume of $O X_{2}$ released at STP, in litres, to two decimal places.

Numerical answer type, answer in L, tolerance 0.01.

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

and

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

. Moles: Na2O2 = 5.85/78 = 0.0750, giving 0.03750 mol O2; KO2 = 6.39/71 = 0.0900, giving 0.06750 mol O2. Total O2 = 0.10500 mol, so volume = 0.10500

\times

22.4 = 2.352 L, i.e. 2.35 L.