Sodium peroxide reacts vigorously with water to produce sodium hydroxide and oxygen.

The unbalanced equation is
Na2O2(s) + H2O(ℓ)→ NaOH(aq) + O2(g). What mass of O2 is produced when 81 g of Na2O2
react?

To determine the mass of O2 produced when 81 g of Na2O2 reacts, we need to balance the equation first. Here's how we can do that:

Na2O2(s) + H2O(ℓ) → NaOH(aq) + O2(g)

There are two Na atoms on the left side and one on the right side, so we need to balance the sodium atoms:

Na2O2(s) + H2O(ℓ) → 2NaOH(aq) + O2(g)

Next, there are two oxygen atoms on the left side and two on the right side, so the equation is balanced.

Now that we have the balanced equation, we can use stoichiometry to determine the mass of O2 produced. The stoichiometric ratio tells us that one mole of Na2O2 reacts to produce one mole of O2.

To calculate the number of moles of Na2O2, we need to determine its molar mass. The molar mass of Na2O2 is:
(2 * atomic mass of Na) + atomic mass of O2
= (2 * 22.99 g/mol) + (2 * 16.00 g/mol)
= 45.98 g/mol + 32.00 g/mol
= 77.98 g/mol

Now we can calculate the number of moles of Na2O2:
moles of Na2O2 = mass of Na2O2 / molar mass of Na2O2
moles of Na2O2 = 81 g / 77.98 g/mol
moles of Na2O2 = 1.038 mol

According to the balanced equation, 1 mole of Na2O2 produces 1 mole of O2. Therefore, the number of moles of O2 produced is also 1.038 mol.

To determine the mass of O2, we can use the molar mass of O2. The molar mass of O2 is 32.00 g/mol.

mass of O2 = moles of O2 * molar mass of O2
mass of O2 = 1.038 mol * 32.00 g/mol
mass of O2 = 33.216 g

Therefore, 33.216 g of O2 is produced when 81 g of Na2O2 reacts.