Answered

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


2Na2O2(s) + 2H2O(l) ---> O2(g) + 4NaOH(aq)

ACalculate the mass of Na2O2 in grams needed to form 4.80g of oxygen.
BHow 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?

Verified Answer :

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:

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.

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:

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

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 = \(\frac{2}[1}\times 0.15=0.3mol\) 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:

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

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 = \(\frac{4}[1}\times 0.15=0.6mol\) 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:

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

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:

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

By Stoichiometry of the reaction:

2 moles of \(Na_2O_2\) produces 1 mole of oxygen gas

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

Putting values in equation 1, we get:

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

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

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