**Reaction**: CCl_{4} + HF → CCl_{2}F_{2 }+ HCl

**For this problem we need to do these steps:**

**S tep 1: Write the balanced equation for the reaction **

**Step 2: Determine moles CCl _{4} reacted using its molar mass and its given mass_{ }**

**Step 3: Calculate the theoretical yield (in g) of CCl_{2}F_{2} stoichiometrically (mole to mole comparison) and its molar mass (mole to mass calculation)**

*Step 4: Calculate percent yield using the actual yield and theoretical yield *

**Step 1: Balanced equation (add coefficients)**

**CCl _{4} + **

*reactants products*

C- 1 C - 1

Cl- 4 Cl - ~~3~~ 4

F- ~~1~~ 2 F - 2

H- ~~1~~ 2 H - ~~1~~ 2

**Balanced equation: CCl _{4} + **

**Step 2: Moles CCl _{4}**

- Molar mass CCl
_{4}= 153.82 g/mol

$\mathbf{32}\mathbf{.}\mathbf{9}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{CCl}}_{\mathbf{4}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{CCl}}_{{\mathbf{4}}}}{\mathbf{153}\mathbf{.}\mathbf{82}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{CCl}}_{\mathbf{4}}}}$**= 0.2139 mol CCl _{4}**

Freon-12, CCl_{2}F_{2}, is prepared from CCl_{4} by reaction with HF. The other product of this reaction is HCl. Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of CCl_{2}F_{2} from 32.9 g of CCl_{4}. Freon-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield.

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Our data indicates that this problem or a close variation was asked in Chemistry - OpenStax 2015th Edition. You can also practice Chemistry - OpenStax 2015th Edition practice problems.