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 24 g of
Na2O2 react?

What mass of water is needed to react completely
with the Na2O2?

Answer 1

#1. Balance the chemical reaction first.
2 Na2O2 + 2 H2O −--> 4 NaOH + O2

#2. Then calculate the molar mass of Na2O2 and O2. From the periodic table, Na - 23 and O - 16. Thus,
Na2O2 : 2*23 + 2*16 = 78 g/mol
O2 : 2*16 = 32 g/mol

#3. Get the moles of Na2O2 from the given 24 g Na2O2, by dividing this mass by the molar mass:
24 g / 78 g/mol = 0.30769 mol Na2O2

#4. From the balanced chemical reaction, we make a mole ratio of O2 : Na2O2. That is, for every mole of O2 produced, there are 2 moles of Na2O2 reacted, or
1 mol O2 / 2 mol Na2O2

#5. Multiply the moles of Na2O2 by the ratio to cancel units:
0.30769 mol Na2O2 * 1 mol O2 / 2 mol Na2O2 = 0.15385 mol O2

#6. Finally, multiply the moles of O2 to its molar mass to get the mass:
0.15385 mol O2 * 32 g/mol = 4.923 g O2


Now, try doing the second question.
Hope this helps~ `u`