Question

2 Na2O2 + 2 H2O → 4NaOH+02
According to the above equation if 1.56g of sodium peroxide reacts with water
calculate
(i) The volume of oxygen gas liberated.
(ii) The mass of sodium hydroxide formed.
[0=16, Na=23, N=14, H=1]​

Answer 1

Answer:

The volume of Oxygen liberated is 224 cm³, whereas the mass of Sodium Hydroxide formed is 1.6 grams

Explanation:

Let's calculate the reacting moles of Sodium peroxide.

Molar mass of Sodium Peroxide = (23 × 2) + (16 × 2) =  78 g/mol

Mass of Sodium peroxide in the solution = 1.56 g

Mole of Sodium Peroxide = 1.56/78 = 0.02 moles.

i) Moles of Oxygen produced.

Mole ratio of Sodium peroxide to Oxygen = 2 : 1

So, moles of Oxygen = 1/2 × 0.02 = 0.01 moles

Molar gas volume at STP is:

1 mole = 22.4 liters

The volume of Oxygen = 0.01 × 22.4 = 0.224 liters

= 224 cm³

ii) Mass of Sodium Hydroxide formed

Mole ratio of Sodium Peroxide to Sodium Hydroxide = 2 : 4 = 1 : 2

Moles of Sodium Hydroxide produced = 0.02 × 2 = 0.04 moles

Molar mass of Sodium Hydroxide = (23 + 16 + 1) = 40 g/mole

Mass of Sodium Hydroxide produced = 40 × 0.04 = 1.6 grams

Answer 2

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