How many grams of fluorine are required to react with 100g of NaBr in the following reaction: F2 + 2NaBr --> 2NaF + Br2
Also, does this show 1 F2 molecule or 2 F2 molecules and how will i write it with the conversion factors with units?
4 answers
Thnx, DrBob for that last answer to my question btw
This is what I have, but teacher said I am showing the fluorine wrong somehow:
F2 + 2NaBr ---> 2NaF + Br2
Moles NaBr = 100g NaBr/102.8938g NaBr/ 1mole NaBr =0.972 moles NaBr
Moles F2 needed = 0.972 moles NaBr/2 molecules F/1mole NaBr =0.486 moles F2
mass F2 = 0.486 moles F2 x 37.9968g F2 /1mole F2 = 18.5 g F2
F2 + 2NaBr ---> 2NaF + Br2
Moles NaBr = 100g NaBr/102.8938g NaBr/ 1mole NaBr =0.972 moles NaBr
Moles F2 needed = 0.972 moles NaBr/2 molecules F/1mole NaBr =0.486 moles F2
mass F2 = 0.486 moles F2 x 37.9968g F2 /1mole F2 = 18.5 g F2
How do i illustrate the Fluorine correctly in relation to the conversion factors??
mols NaBr = 100 g NaBr x (1 mol NaBr/102.89) = 0.972 mols NaBr.
0.972 mols NaBr x (1 mol F2/2 mols NaBr) = 0.486 mols F2.
g F2 = 0.486 mols F2 x (19 g F2/1 mol) = ?
I think this step in your work is what your teacher would prefer you did differently.
Moles F2 needed = 0.972 moles NaBr/2 molecules F/1mole NaBr =0.486 moles F2
You have two points of contention.
1. You have written in an extra step that isn't necessary; i.e. you have converted (actually TRIED to convert) mols to molecules and back to moles) and that isn't necessary. It only leads to confusion.
2. Your factor (0.972 mols NaBr/2 molecules F2) is not right. 0.972 mols NaBr/2 atoms F would be ok but then you need to convert from atoms to mols. That's an extra step that isn't needed. I THINK you may be confusing molecules with mols. A single molecule is an atom of F + an atom of F. A mol F2 is not a single molecule (two atoms) but 6.02E23
molecules and that's a huge difference. That 6.02E23 molecules (1 mol F2) has a mass of 19 grams.
Follow up if this isn't clear. I have shown how to work the problem above.
0.972 mols NaBr x (1 mol F2/2 mols NaBr) = 0.486 mols F2.
g F2 = 0.486 mols F2 x (19 g F2/1 mol) = ?
I think this step in your work is what your teacher would prefer you did differently.
Moles F2 needed = 0.972 moles NaBr/2 molecules F/1mole NaBr =0.486 moles F2
You have two points of contention.
1. You have written in an extra step that isn't necessary; i.e. you have converted (actually TRIED to convert) mols to molecules and back to moles) and that isn't necessary. It only leads to confusion.
2. Your factor (0.972 mols NaBr/2 molecules F2) is not right. 0.972 mols NaBr/2 atoms F would be ok but then you need to convert from atoms to mols. That's an extra step that isn't needed. I THINK you may be confusing molecules with mols. A single molecule is an atom of F + an atom of F. A mol F2 is not a single molecule (two atoms) but 6.02E23
molecules and that's a huge difference. That 6.02E23 molecules (1 mol F2) has a mass of 19 grams.
Follow up if this isn't clear. I have shown how to work the problem above.
0 answers