What mass of oxygen is liberated?

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In summary, the problem is asking for the mass of oxygen gas that is released when a 2.5 gram sample of sodium nitrate is heated. Using the given information and the equation 2NaNO3 ---> 2NaNO2 + O2, the solution is found to be 0.47 grams of O2. The 85 grams of NaNO3 and 32 grams of O2 used in the calculation are related to the atomic mass of each element. A table of atomic masses can be found in the book. Visualizing the atoms and molecules as blobs may also help in understanding the concept.
  • #1
science_rules
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Homework Statement


This is a solved example problem in my chemistry book: What mass of oxygen gas O2 is liberated when a 2.5 gram sample of sodium nitrate NaNO3 is heated? Please keep in mind that I am self-studying, I am not currently in chemistry class.

Homework Equations


2NaNO3 ---> 2NaNO2 + O2
Known: 2.5 g NaNO3, 85 g of NaNO3/1 mole NaNO3, 1 mole O2/2 moles NaNO3, and 32 g O2/1mole O2
Unknown: mass O2 g

The Attempt at a Solution


This is how it was solved: 2.5 g NaNO3 X (1 mole 3/85 g NaNO3) X (1 mole O2/2 moles NaNO3) X (32 g O2/1 mole O2) = 0.47 g O2 What I don't understand is how they got the 85 grams of NaNO3 and the 32 grams of O2? Can someone please show me step-by-step how they got the 85 grams and the 32 grams?? Are these related to the atomic mass of each element?
 
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  • #2
science_rules said:

Homework Statement


This is a solved example problem in my chemistry book: What mass of oxygen gas O2 is liberated when a 2.5 gram sample of sodium nitrate NaNO3 is heated? Please keep in mind that I am self-studying, I am not currently in chemistry class.

Homework Equations


2NaNO3 ---> 2NaNO2 + O2
Known: 2.5 g NaNO3, 85 g of NaNO3/1 mole NaNO3, 1 mole O2/2 moles NaNO3, and 32 g O2/1mole O2
Unknown: mass O2 g

The Attempt at a Solution


This is how it was solved: 2.5 g NaNO3 X (1 mole 3/85 g NaNO3) X (1 mole O2/2 moles NaNO3) X (32 g O2/1 mole O2) = 0.47 g O2 What I don't understand is how they got the 85 grams of NaNO3 and the 32 grams of O2? Can someone please show me step-by-step how they got the 85 grams and the 32 grams?? Are these related to the atomic mass of each element?

Yes they are. There is surely a table of atomic masses in your book. And these are roughly 23 for Na, 14 for N and 16 for O. 23 + 14 + 3×16 = 85. Likewise 2×13 = 32.

Why they do these calculations in this way is surely explained in your book.

We can't substitute for your book - I would only say that if you have difficulty then with the formulae then imagine the atoms as little blobs and draw them so the molecule has say a black blob for Na, a gray blob for N and three white blobs for the three O in NaNO3. Each blob has the weight or mass that we said. Then in a chemical reaction you picture them as moving about. Maybe this will make it clear for you at the start and later you will be able to do without it

The guy who invented the atomic theory represented atoms and molecules this way, but it is too cumbersome for printers.
 
  • #3
okay, thankyou, I will look at my book.
 

1. What is the chemical equation for the liberation of oxygen?

The chemical equation for the liberation of oxygen is 2H2O → 2H2 + O2, where water (H2O) is broken down into hydrogen gas (H2) and oxygen gas (O2).

2. How is oxygen liberated in the laboratory?

Oxygen can be liberated in the laboratory by heating certain compounds that contain oxygen, such as potassium chlorate (KClO3). When heated, potassium chlorate decomposes into potassium chloride (KCl) and oxygen gas (O2).

3. What factors affect the amount of oxygen liberated?

The amount of oxygen liberated can be affected by the amount of starting material, the efficiency of the reaction, and the temperature and pressure conditions. It can also be affected by impurities or contaminants in the starting material.

4. How is the mass of oxygen liberated calculated?

The mass of oxygen liberated can be calculated using the ideal gas law, which states that the amount of gas (in moles) is proportional to its volume, pressure, and temperature. By knowing the volume and pressure of the liberated oxygen, as well as the temperature, the mass can be calculated using the equation m = PV/RT, where m is the mass, P is the pressure, V is the volume, R is the gas constant, and T is the temperature.

5. Why is the liberation of oxygen important in biology?

Oxygen is essential for many biological processes, such as cellular respiration, which provides energy for the body. The liberation of oxygen allows organisms to use oxygen to break down nutrients and produce energy. It is also important in photosynthesis, where plants use sunlight to convert carbon dioxide and water into oxygen and glucose.

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