How many grams of oxygen gas will be produced if 289 g of NaClO3 is decomposed to sodium chloride and oxygen gas? (the fourth letter in this formula is not capital i, but a lower case l)
2NaClO3 ==> 2NaCl + 3O2
This is a simple stoichiometry problem. I went over this in detail for Veronica. Here is a link to that question. Post your work if you get stuck.
https://www.jiskha.com/questions/1815058/i-cant-get-this-if-4-00-g-of-h2-are-made-to-react-with-excess-co-g-how-many-grams-of
To determine the number of grams of oxygen gas produced, we need to know the balanced chemical equation for the decomposition of sodium chlorate (NaClO₃) to sodium chloride (NaCl) and oxygen gas (O₂).
The balanced equation is:
2 NaClO₃ → 2 NaCl + 3 O₂
From the balanced equation, we can see that 2 moles of NaClO₃ decompose to produce 3 moles of O₂.
Now, we need to calculate the number of moles of NaClO₃ present in 289 g. To do this, we'll use the molar mass of NaClO₃.
Molar mass of NaClO₃:
Na: 22.99 g/mol
Cl: 35.45 g/mol
O: 16.00 g/mol
Molar mass of NaClO₃ = 22.99 g/mol + 35.45 g/mol + 3(16.00 g/mol) = 122.44 g/mol
Now, we'll calculate the number of moles of NaClO₃:
Moles of NaClO₃ = Mass of NaClO₃ / Molar mass of NaClO₃
= 289 g / 122.44 g/mol
≈ 2.36 mol
Since the molar ratio between NaClO₃ and O₂ is 2:3, we can calculate the moles of O₂ produced:
Moles of O₂ = (3/2) * Moles of NaClO₃
= (3/2) * 2.36 mol
≈ 3.54 mol
Finally, we'll convert the moles of O₂ to grams:
Grams of O₂ = Moles of O₂ * Molar mass of O₂
= 3.54 mol * 32.00 g/mol (molar mass of O₂)
≈ 113.3 g
Therefore, approximately 113.3 grams of oxygen gas will be produced when 289 grams of NaClO₃ is decomposed.