If you reacted 100.0 g of carbon tetrachloride (CCl4) with excess antimony fluoride (SbF3), how many grams of carbon dichloride difluoride would be produced?

Do you have an equation for the reaction?

I assume the reaction is
3CCl4 + 2SbF3 ==> 3CCl2F2 + 2SbCl3

mols CCl4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols CCl4 to mols CCl2F2.
Then g CCl2F2 = mols CCl2F2 x molar mass CCl2F2 = ?

To determine the number of grams of carbon dichloride difluoride (CCl2F2) produced, we need to calculate the theoretical yield of the reaction.

To do this, we must first write and balance the chemical equation for the reaction between carbon tetrachloride (CCl4) and antimony fluoride (SbF3):

CCl4 + SbF3 -> CCl2F2 + SbCl3

The balanced equation tells us that 1 mol of CCl4 reacts with 1 mol of SbF3 to produce 1 mol of CCl2F2 and 1 mol of SbCl3.

Next, we need to determine the molar mass of carbon tetrachloride (CCl4) and carbon dichloride difluoride (CCl2F2):

- Molar mass of CCl4 = (1 atom of C * atomic mass of C) + (4 atoms of Cl * atomic mass of Cl)
- Molar mass of CCl2F2 = (1 atom of C * atomic mass of C) + (2 atoms of Cl * atomic mass of Cl) + (2 atoms of F * atomic mass of F)

Using the atomic masses from the periodic table:

- Molar mass of CCl4 = (1 * 12.01) + (4 * 35.45) = 153.82 g/mol
- Molar mass of CCl2F2 = (1 * 12.01) + (2 * 35.45) + (2 * 18.99) = 131.01 g/mol

Now we can calculate the theoretical yield of CCl2F2 using stoichiometry:

1. Convert the given mass of CCl4 to moles:
- Moles of CCl4 = (Given mass of CCl4) / (Molar mass of CCl4)
- Moles of CCl4 = 100.0 g / 153.82 g/mol = 0.65 mol

2. Use the mole ratio from the balanced equation to determine moles of CCl2F2:
- Moles of CCl2F2 = (Moles of CCl4) * (Mole ratio of CCl2F2 / CCl4) = 0.65 mol * (1 mol CCl2F2 / 1 mol CCl4) = 0.65 mol

3. Convert moles of CCl2F2 to grams:
- Mass of CCl2F2 = (Moles of CCl2F2) * (Molar mass of CCl2F2)
- Mass of CCl2F2 = 0.65 mol * 131.01 g/mol = 85.16 g

Therefore, the theoretical yield of carbon dichloride difluoride (CCl2F2) from the given reaction is approximately 85.16 grams.