Question

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?

Answer 1

First, we balance the chemical reaction:
2 Na2O2 + 2 H2O → 4 NaOH + O2

Then, we need to determine the molar masses of Na2O2 and O2 in order to convert them to moles. To get them, we need the mass of each element in the formula and add. In the periodic table, Na = 23 and O = 16. Thus the respective molar masses are:
Na2O2: 2*23 + 2*16 = 78 g/mol
O2: 2*16 = 32 g/mol

Then we use the given, 81 g of Na2O2. To get the moles of Na2O2, we divide the mass by the molar mass:
81 g / (78 g/mol) = 1.03846 moles Na2O2

From the balanced reaction, 1 mole of O2 is being produced for every 2 moles of Na2O2 reacted. Thus,
1.03846 moles Na2O2 * (1 mole O2 / 2 moles Na2O2) = 0.5192 moles O2

Finally, we multiply this by the molar mass of O2 to get its mass:
0.5192 * 32 = 16.62 g O2

Hope this helps~ `u`

Answer 2

Thank YOU!!!!