# Calculate the volume of oxygen gas, measured at 25 C and pressure of 101.3 kPa, that is available from the reaction of 0.5 kg of lithium peroxide with carbon dioxide according to the...

Calculate the **volume** of oxygen gas, measured at 25 C and pressure of 101.3 kPa, that is available from the reaction of 0.5 kg of lithium peroxide with carbon dioxide according to the equation:

2Li2O2 (s) + 2CO2 (g) ---> 2Li2CO3 (s) + O2 (g)

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The equation between lithium peroxide and carbon dioxide is: 2Li2O2 (s) + 2CO2 (g) ---> 2Li2CO3 (s) + O2 (g)

Two moles of lithium peroxide react to release one mole of oxygen. The molar mass of lithium peroxide is 45.881 g/mol. 0.5 kg of lithium peroxide is equivalent to 10.89 moles.

The temperature is given as 25 C which is equal to 298 K and the pressure is 101.3 kPa. Using the ideal gas law, the volume of oxygen released can be estimated. The relation between volume V, pressure P, number of moles of the gas n, and temperature T is PV = n*RT where R is a constant equal to 8.314 L*kPa/K*mol

Substituting the values we have

V = 10.89*8.314*298/101.3*2

=> 133.17 L

The volume occupied by oxygen released when the given reaction takes place at the given pressure and temperature is 133.17 L