Oxygen is produced by the reaction of sodium peroxide and water.

2na2o2(s) + 2h2o(l) > o2(g) + 4naoh(aq)

a. calculate the mass of na2o2 in grams needed to form 4.80g of oxygen.
b. how many grams of naoh are produced when 4.80 g of o2 is formed?
c. when 0.48g of na2o2 is dropped in water, how many grams of o2 are formed?

For a: The mass of required is 23.4 grams

For b: The mass of NaOH required is 24 grams

For c: The mass of oxygen gas produced is 0.0984 grams.

Explanation:

To calculate the number of moles, we use the equation:

    .....(2)

Given mass of oxygen gas = 4.80 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 2, we get:

For the given chemical equation:

By Stoichiometry of the reaction:

1 mole of oxygen gas is produced from 2 moles of

So, 0.15 moles of oxygen gas will be produced from = of

Now, calculating the mass of by using equation 1:

Molar mass of = 78 g/mol

Moles of = 0.3 moles

Putting values in equation 1, we get:

Hence, the mass of required is 23.4 grams

By Stoichiometry of the reaction:

When 1 mole of oxygen gas is produced, then 4 moles of NaOH is also produced.

So, when 0.15 moles of oxygen gas will be produced, then = of NaOH will be produced.

Now, calculating the mass of NaOH by using equation 1:

Molar mass of NaOH = 40 g/mol

Moles of NaOH = 0.6 moles

Putting values in equation 1, we get:

Hence, the mass of NaOH required is 24 grams

Given mass of = 0.48 g

Molar mass of = 78 g/mol

Putting values in equation 2, we get:

By Stoichiometry of the reaction:

2 moles of produces 1 mole of oxygen gas

So, moles of will produce = of oxygen gas

Now, calculating the mass of oxygen gas by using equation 1:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = moles

Putting values in equation 1, we get:

Hence, the mass of oxygen gas produced is 0.0984 grams.