The value of ΔH° for the reaction below is -126 kJ. The amount of heat that is released by the reaction of 40.0 g of Na2O2 with water is kJ. 2Na2O2 (s) + 2H2O (l) → 4NaOH (s) + O2 (g)

-32.3 kJ

Explanation:

Step 1: Write the balanced thermochemical equation

2 Na₂O₂(s) + 2 H₂O(l) ⇒ 4 NaOH(s) + O₂(g)   ΔH° = -126 kJ

Step 2: Calculate the moles corresponding to 40.0 g of Na₂O₂

The molar mass of Na₂O₂ is 77.98 g/mol.

40.0 g × 1 mol/77.98 g = 0.513 mol

Step 3: Calculate the amount of heat released upon the reaction of 0.513 moles of Na₂O₂

According to the thermochemical equation, 126 kJ of heat are released every 2 moles of Na₂O₂.

0.513 mol Na₂O₂ × (-126 kJ/2 mol Na₂O₂) = -32.3 kJ

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.