Question

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?

Answer 1

first you should start by calculating the molar mass of each compounds and elements so molar mass for Na2O2 = 77.98g, O2=16.0g and NaOH= 39.997g now we start with setting up the problems

a) 4.80g O2 (1mol O2/16g o2)(2 mol Na2O2/1mol O2)(77.98g NaO2/zmol NaO2) =46.8 g of Na2O2 are required to form 4.80g of oxygen.

b) 4.80g O2 (1mol O2/16g O2) (4 mol NaOH/1 mol O2) (33.997g naOH/1 mol NaOH)= 48.0 g are produced when 4.80g of O2 is formed

c) 0.48g Na2O2 (1mol Na2O2/77.98g Na2O2) (1 mol O2/2 mol Na2O2) (16.0g O2/ 1 mol O2)= 0.049g of O2 are formed.

Answer 2

For a: The mass of
`Na_2O_2` 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.

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(2)

• For oxygen gas:

Given mass of oxygen gas = 4.80 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 2, we get:

`\text{Moles of oxygen gas}=(4.80g)/(32g/mol)=0.15mol`

For the given chemical equation:

`2Na_2O_2(s)+2H_2O(l)\rightarrow O_2(g)+4NaOH(aq.)`

• For a:

By Stoichiometry of the reaction:

1 mole of oxygen gas is produced from 2 moles of
`Na_2O_2`

So, 0.15 moles of oxygen gas will be produced from =
`\frac{2}[1}* 0.15=0.3mol` of
`Na_2O_2`

Now, calculating the mass of
`Na_2O_2` by using equation 1:

Molar mass of
`Na_2O_2` = 78 g/mol

Moles of
`Na_2O_2` = 0.3 moles

Putting values in equation 1, we get:

`0.3mol=\frac{\text{Mass of }Na_2O_2}{78g/mol}\\\\\text{Mass of }Na_2O_2=(0.3mol* 78g/mol)=23.4g`

Hence, the mass of
`Na_2O_2` required is 23.4 grams

• For b:

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 =
`\frac{4}[1}* 0.15=0.6mol` 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:

`0.6mol=\frac{\text{Mass of NaOH}}{40g/mol}\\\\\text{Mass of NaOH}=(0.6mol* 40g/mol)=24g`

Hence, the mass of NaOH required is 24 grams

• For c:

Given mass of
`Na_2O_2` = 0.48 g

Molar mass of
`Na_2O_2` = 78 g/mol

Putting values in equation 2, we get:

`\text{Moles of }Na_2O_2=(0.48g)/(78g/mol)=6.15* 10^(-3)mol`

By Stoichiometry of the reaction:

2 moles of
`Na_2O_2` produces 1 mole of oxygen gas

So,
`6.15* 10^(-3)` moles of
`Na_2O_2` will produce =
`\frac{1}[2}* 6.15* 10^(-3)=3.075mol` 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 =
`3.075* 10^(-3)` moles

Putting values in equation 1, we get:

`3.075* 10^(-3)mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(3.075* 10^(-3)mol* 32g/mol)=0.0984g`

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