Read the given equation. Na2O2 + CO2 → Na2CO3 + O2 What volume of O2 gas is produced from 2.80 liters of CO2 at STP?

1.40 L.

Explanation:

It is a stichiometric problem. Firstly, we should wright the reaction as a balanced equation:

Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂

It is clear that 1.0 mole of Na₂O₂ reacts with 1.0 mole of CO₂ to produce 1.0 mole of Na₂CO₃ and 0.5 mole of O₂. Then we should convert the volume of CO₂ (2.80 L) to moles via using the gas law of ideal gas: PV = nRT, n = PV / RT,

Where, P is the pressure of the gas in atm (P at STP = 1.0 atm).

V is the volume of the gas in L (V = 2.80 L).

R is the general gas constant (R = 0.082 L. atm/mol. K).

T is the temperature in K (T at STP = 0.0 °C + 273 = 273 K).

∴ n of CO₂ = PV / RT = (1.0 atm) (2.80 L) / (0.082 L. atm/mol. K) (273 K) = 2.80 / 22.386 = 0.125 mole.

From the stichiometry:

1.0 mole of CO₂ will produce → 0.50 mole of O₂

0.125 mole of CO₂ will produce →? mole of O₂

The number of moles of O₂ produced = (0.125 mole of CO₂) (0.50 mole of O₂) / (1.0 mole of CO₂) = 0.06254 mole. Finally, we can convert the number of moles of O₂ produced to volume using the mentioned gas law of ideal gas: V = nRT/P. V of O₂ produced = nRT / P = (0.06254 mole) (0.082 L. atm/mol. K) (273.0 K) / (1.0 atm) = 1.40 L.