Freon-12 is formed as follows:

3 CCl4 + 2 SbF3 -> 3 CCl2F2 + 2 SbCl3

150 g tetra chloromethane is combined with 100 g antimony (III) fluoride to give Freon-12 (CCl2F2).

Identify the limiting and excess reagents, and calculate how many grams of Freon-12 can be formed.

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  1. mols CCl4 = grams/molar mass = ?
    mols SbF3 = grams/molar mass = ?

    Convert mols CCl4 to mols SbF3. If the SbF3 mols is greater than CCl4 is the limiting reagent. If not SbF3 is the LR.
    Post your work if you get stuck.

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  2. I got SbF3 as the limiting reagent.

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  3. I agree

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