2. Sodium peroxide reacts vigorously with water in the following unbalanced equation: Na_(2)O_(2)(s)+H_(2)O(l)arrow NaOH(aq)+O_(2)(g) a. What mass of O_(2) is produced when 50.0g of Na_(2)O_(2) react with H_(2)O b. What mass in grams of water is needed to react completely with Na_(2)O_(2)

a. To find the mass of O2 produced when 50.0g of Na2O2 react with H2O, we first need to balance the chemical equation:<br /><br />2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g)<br /><br />Now, we can use stoichiometry to find the mass of O2 produced. The molar mass of Na2O2 is 77.98 g/mol, and the molar mass of O2 is 32.00 g/mol.<br /><br />First, we need to convert the mass of Na2O2 to moles:<br />50.0 g Na2O2 * (1 mol Na2O2 / 77.98 g Na2O2) = 0.643 mol Na2O2<br /><br />According to the balanced equation, 2 moles of Na2O2 produce 1 mole of O2. So, we can find the moles of O2 produced:<br />0.643 mol Na2O2 * (1 mol O2 / 2 mol Na2O2) = 0.3215 mol O2<br /><br />Finally, we can convert the moles of O2 to grams:<br />0.3215 mol O2 * (32.00 g O2 / 1 mol O2) = 10.30 g O2<br /><br />Therefore, 10.30 grams of O2 are produced when 50.0g of Na2O2 react with H2O.<br /><br />b. To find the mass of water needed to react completely with Na2O2, we can use stoichiometry again. The molar mass of H2O is 18.02 g/mol.<br /><br />According to the balanced equation, 2 moles of Na2O2 react with 2 moles of H2O. So, the moles of H2O needed are the same as the moles of Na2O2:<br />0.643 mol Na2O2 * (2 mol H2O / 2 mol Na2O2) = 0.643 mol H2O<br /><br />Now, we can convert the moles of H2O to grams:<br />0.643 mol H2O * (18.02 g H2O / 1 mol H2O) = 11.56 g H2O<br /><br />Therefore, 11.56 grams of water are needed to react completely with 50.0g of Na2O2.