When it comes to understanding the amount of water required for a reaction involving 125.0 g Na2O2, it is important to understand the molar mass of Na2O2 and how much water will be needed for that specific reaction. The molar mass of Na2O2 is 61.99 g/mol which means that 125.0 grams of Na 2 O 2 would equal approximately 2 moles (125.0 / 61.99 = 1.998).

For this type of reaction, we need to use the following equation in order to determine how many grams (and milliliters) are necessary: 2Na2O + 4H20 –>4NaOH + O2 . In order for two moles of sodium oxide to react with an adequate amount of water, four moles of H20 must be present in order for there to be enough molecules available during the reaction process; this can be written out as 4 x 18g/mol = 72 grams (or mL) of H20 needed per mole of sodium oxide when performing this type of chemical reaction – resulting in 144 gram (or mL) total being required when using two moles worth 125g or Na2O3.