5. Sodium peroxide reacts vigorously with water in the following unbalanced equation:
Na202 + H20+ NaOH + O2
What mass of O2 is produced when 50.0 g Na2O2 reacts with H2O?
Show your work

30.7 g of O₂ are produced

Explanation:

This is a redox reaction, where the peroxide is reduced to oxygen and oxidized to hydroxide. The equation is this one:

2Na₂O₂ + 2H₂O → 4NaOH + 3O₂

We assume that water is in excess, so the peroxide is the limiting reagent.

We convert the mass to moles: 50 g . 1mol / 78 g = 0.641 moles

2 moles of peroxide can produce 3 moles of oxygen (according to stoichiometry)

Then, 0.641 moles of peroxide will produce (0.641 . 3 ) /2 = 0.961 moles O₂

Finally, we convert the moles to mass: 0.961mol . 32 g / 1mol = 30.7 g of O₂

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