Problem: Freon-12, CCl2F2, is prepared from CCl4 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 CCl2F2 from 32.9 g of CCl4. 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.

Reaction: CCl₄ + HF → CCl₂F₂ + HCl 

For this problem we need to do these steps:

Step 1: Write the balanced equation for the reaction 

Step 2: Determine moles CCl₄ reacted using its molar mass and its given mass 

Step 3: Calculate the theoretical yield (in g) of CCl₂F₂ 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)

CCl₄ + 2HF → CCl₂F₂ + 2HCl 

reactants               products

C- 1                   C - 1

Cl- 4                  Cl - 3 4  

F- 1 2                 F - 2

H- 1 2                 H - 1 2  

Balanced equation: CCl₄ + 2HF → CCl₂F₂ + 2HCl 

Step 2: Moles CCl₄

• Molar mass CCl₄ = 153.82 g/mol

$\mathbf{32}\mathbf{.}\mathbf{9}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{CCl}}_{\mathbf{4}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{CCl}}_{\mathbf{4}}}{\mathbf{153}\mathbf{.}\mathbf{82}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{CCl}}_{\mathbf{4}}}}$= 0.2139 mol CCl₄