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?
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
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.
To determine the amount of fluorine (F2) needed to react with 100g of NaBr, you can use stoichiometry and molar ratios.
First, let's find the molar mass of NaBr and F2:
- The molar mass of NaBr (sodium bromide) is calculated as:
Na = 22.99 g/mol
Br = 79.90 g/mol
Molar mass of NaBr = 22.99 + 79.90 = 102.89 g/mol
- The molar mass of F2 (fluorine gas) is calculated as:
F = 19.00 g/mol
Molar mass of F2 = 2 * 19.00 = 38.00 g/mol
Now, let's set up the stoichiometric ratio using the balanced equation:
F2 + 2NaBr -> 2NaF + Br2
From the balanced equation, we can see that it takes 1 mole of F2 to react with 2 moles of NaBr.
To calculate the amount of F2 needed, follow these steps:
Step 1: Convert the given mass of NaBr to moles using its molar mass:
Moles of NaBr = 100g / 102.89 g/mol = 0.971 moles (rounded to 3 decimal places)
Step 2: Use the mole ratio from the balanced equation to determine the moles of F2 required:
Moles of F2 = 0.971 moles of NaBr * (1 mole F2 / 2 moles NaBr)
= 0.4855 moles of F2 (rounded to 4 decimal places)
Step 3: Convert the moles of F2 to grams using its molar mass:
Mass of F2 = 0.4855 moles * 38.00 g/mol = 18.471g (rounded to 3 decimal places)
So, approximately 18.471 grams of F2 are required to react with 100 grams of NaBr.
Now, to address your second question, the coefficient in front of F2 in the balanced equation indicates the number of F2 molecules. In this case, since the coefficient is 1, it represents "1 F2" molecule.
To write this with conversion factors and units, you would have:
1 mole NaBr → 1/2 mole F2 → 38.00 g F2
I hope this helps!