Question

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 prcent yield of a reaction that produces 12.5 g of CCl2F2 from 32.9 g of CCl4. Feron-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone an d has a very long lifetime in the atmosphere. Determine the percent Yield.

Answer 1

Answer:

percent yield = 40.6 %

Explanation:

The question asks to determine the percent yield, which can be defined as:

percent yield = $\frac{actual yield}{theoretical yield} *100$

where the actual yield is how much product was obtained, in this case 12.5 g of CCl₂F₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically.

So we know already the actual yield, we need to calculate the theoretical yield.

First we need to write the reaction chemical equation:

CCl₄ + HF → CCl₂F₂ + HCl

and balance the equation:

CCl₄ + 2 HF → CCl₂F₂ + 2 HCl

In the equation we can see that for every mol of CCl₄ we should get 1 mol of CCl₂F₂ (molar ratio 1:1). So if we calculate the moles of CCl₄ in the given 39.2 g of CCl₄ we could know how many moles of CCl₂F₂ (assuming HF is in excess).

• Moles of CCl₄ = mass CCl₄ / molar mass CCl₄
• Molar Mass CCl₄ = 12.011 + 4 * 35.45 = 153.811 g/mol
• Moles of CCl₄ = 39.2 g / 153.811 g/mol = 0.2549 moles

From the molar ratio we know:

Moles of CCl₂F₂ = moles of CCl₄ = 0.2549 moles

Now we need to convert these moles into grams to get the theoretical yield of CCl₂F₂ in grams:

• mass CCl₂F₂ = moles CCl₂F₂ * molar mass CCl₂F₂
• Molar Mass CCl₂F₂ = 12.011 + 2 * 35.45 + 2 * 18.998 = 120.907 g/mol
• Mass CCl₂F₂ = 0.2549 moles * 120.907 g/mol = 30.81 g

• Theoretical yield CCl₂F₂ = 30.81 g

Percent yield = (12.5 g / 30.81 g) * 100 = 40.6 %