What Mass of Oxygen Gas Is Liberated from Heating Sodium Nitrate?

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SUMMARY

The discussion centers on calculating the mass of oxygen gas (O2) liberated from heating a 2.5 gram sample of sodium nitrate (NaNO3). The balanced chemical equation is 2NaNO3 ---> 2NaNO2 + O2. The calculation involves using the molar mass of NaNO3, which is 85 g/mol, and the molar mass of O2, which is 32 g/mol. The final result of the calculation shows that 0.47 grams of O2 are produced from the reaction.

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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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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.
 
okay, thankyou, I will look at my book.
 

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