Question

When \(4.0 \mathrm{~mol}\) of \(\mathrm{CCl}_{4}\) reacts with an excess of \(\mathrm{HF}, 3.0 \mathrm{~mol}\) of \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) (Freon) is obtained. The equation for the reaction is $$ \mathrm{CCl}_{4}(l)+2 \mathrm{HF}(g) \longrightarrow \mathrm{CCl}_{2} \mathrm{~F}_{2}(l)+2 \mathrm{HCl}(g) $$ State which of the statements are true about the reaction and make the false statements true. (a) The theoretical yield for \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is \(3.0 \mathrm{~mol}\). (b) The theoretical yield for \(\mathrm{HCl}\) is \(71 \mathrm{~g}\). (c) The percent yield for the reaction is \(75 \%\). (d) The theoretical yield cannot be determined unless the exact amount of \(\mathrm{HF}\) is given. (e) From just the information given above, it is impossible to calculate how much HF is unreacted. (f) For this reaction, as well as for any other reaction, the total number of moles of reactants is equal to the total number of moles of product. (g) Half a mole of \(\mathrm{HF}\) is consumed for every mole of \(\mathrm{CCl}_{4}\) used. (h) At the end of the reaction, no \(\mathrm{CCl}_{4}\) is theoretically left unreacted.

Short answer

Statement (a): False. The correct statement should be: The theoretical yield for \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is \(4.0 \mathrm{~mol}\). Statement (b): False. The correct statement should be: The theoretical yield for \(\mathrm{HCl}\) is \(292 \mathrm{~g}\). Statement (c): True. The percent yield for the reaction is \(75 \%\). Statement (d): False. The correct statement should be: The theoretical yield can be determined even if the exact amount of \(\mathrm{HF}\) is not given, as long as the reaction follows the given balanced chemical equation and \(\mathrm{HF}\) is in excess. Statement (e): True. From just the information given above, it is impossible to calculate how much HF is unreacted. Statement (f): False. The law of conservation of mass does not state that the moles of reactants must equal the moles of products. Statement (g): False. The correct statement should be: Two moles of \(\mathrm{HF}\) are consumed for every mole of \(\mathrm{CCl}_{4}\) used. Statement (h): True. At the end of the reaction, no \(\mathrm{CCl}_{4}\) is theoretically left unreacted.

Step-by-step solution

Statement (a)

The theoretical yield for \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is \(3.0 \mathrm{~mol}\). To find the theoretical yield, we need to use the stoichiometry of the reaction. According to the balanced chemical equation, one mole of \(\mathrm{CCl}_{4}\) reacts with two moles of \(\mathrm{HF}\) to yield one mole of \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\). Therefore, if \(4.0 \mathrm{~mol}\) of \(\mathrm{CCl}_{4}\) reacts, the theoretical yield for \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is also \(4.0 \mathrm{~mol}\). So, this statement is false. The correct statement should be: The theoretical yield for \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is \(4.0 \mathrm{~mol}\).

Statement (b)

The theoretical yield for \(\mathrm{HCl}\) is \(71 \mathrm{~g}\). From the balanced chemical equation, we can see that two moles of \(\mathrm{HCl}\) are produced for every mole of \(\mathrm{CCl}_{4}\) used. Thus, if \(4.0 \mathrm{~mol}\) of \(\mathrm{CCl}_{4}\) reacts, the theoretical yield for \(\mathrm{HCl}\) is \(2 \times 4.0 = 8.0 \mathrm{~mol}\). Now, let's convert moles of \(\mathrm{HCl}\) to grams, given that the molar mass of \(\mathrm{HCl}\) is \(36.5 \mathrm{~g/mol}\): $$8.0 \mathrm{~mol} \times 36.5 \cfrac{\mathrm{~g}}{\mathrm{mol}} = 292 \mathrm{~g}$$ The statement is false. The correct statement should be: The theoretical yield for \(\mathrm{HCl}\) is \(292 \mathrm{~g}\).

Statement (c)

The percent yield for the reaction is \(75 \%\). Percent yield is calculated using the formula: $$\text{Percent Yield} = \cfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$$ In this case, the actual yield of \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is \(3.0 \mathrm{~mol}\) and the theoretical yield is \(4.0 \mathrm{~mol}\). Therefore, the percent yield is: $$\text{Percent Yield} = \cfrac{3.0}{4.0} \times 100 = 75 \%$$ This statement is true.

Statement (d)

The theoretical yield cannot be determined unless the exact amount of \(\mathrm{HF}\) is given. As we already determined the theoretical yield for \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) and \(\mathrm{HCl}\) based on the stoichiometry of the reaction and the given amount of \(\mathrm{CCl}_{4}\), this statement is false. The correct statement should be: The theoretical yield can be determined even if the exact amount of \(\mathrm{HF}\) is not given, as long as the reaction follows the given balanced chemical equation and \(\mathrm{HF}\) is in excess.

Statement (e)

From just the information given above, it is impossible to calculate how much HF is unreacted. Since we know that \(4.0 \mathrm{~mol}\) of \(\mathrm{CCl}_{4}\) reacts and the reaction consumes two moles of \(\mathrm{HF}\) for every mole of \(\mathrm{CCl}_{4}\), we can calculate the amount of \(\mathrm{HF}\) consumed: $$4.0 \mathrm{~mol} \times 2 = 8.0 \mathrm{~mol}$$ However, without knowing the initial amount of \(\mathrm{HF}\), we cannot determine how much \(\mathrm{HF}\) is unreacted. Thus, this statement is true.

Statement (f)

For this reaction, as well as for any other reaction, the total number of moles of reactants is equal to the total number of moles of product. This statement does not apply to every reaction. The law of conservation of mass states that the total mass of reactants equals the total mass of products, but it does not state that the moles of reactants must equal the moles of products. In fact, stoichiometry often involves different numbers of moles of reactants and products. So this statement is false.

Statement (g)

Half a mole of \(\mathrm{HF}\) is consumed for every mole of \(\mathrm{CCl}_{4}\) used. According to the balanced chemical equation, two moles of \(\mathrm{HF}\) are consumed for every mole of \(\mathrm{CCl}_{4}\) used. Therefore, this statement is false. The correct statement should be: Two moles of \(\mathrm{HF}\) are consumed for every mole of \(\mathrm{CCl}_{4}\) used.

Statement (h)

At the end of the reaction, no \(\mathrm{CCl}_{4}\) is theoretically left unreacted. Since \(\mathrm{HF}\) is present in excess and the reaction proceeds to completion, the limiting reagent is \(\mathrm{CCl}_{4}\). All the \(\mathrm{CCl}_{4}\) will react to yield \(\mathrm{CCl}_{2}\mathrm{~F}_{2}\) and \(\mathrm{HCl}\). Thus, this statement is true.

Key concepts

Theoretical Yield Calculation

Understanding the concept of theoretical yield is fundamental in the study of chemical reactions. The theoretical yield is the maximum amount of product that can be generated as predicted by the stoichiometry of the reaction, assuming perfect conditions where 100% of the reactants are converted to products. To calculate it, we must start by looking at the balanced chemical equation. For instance, when we react \(4.0 \)mol of \(\mathrm{CCl}_{4}\) with an excess of HF, a balanced equation indicates the mole ratio between \(\mathrm{CCl}_{4}\) and \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is 1:1.

Therefore, we anticipate a theoretical yield of \(4.0 \)mol for \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\). It's important for students to identify the reactants and products in a reaction equation and use the coefficients to relate amounts accurately. Exercises often provide the amount of only one reactant, and by comprehending the concept of theoretical yield, students can determine how much of a product could theoretically be formed without knowing the exact quantities of all reactants involved.

Percent Yield Formula

Once we understand theoretical yield, the next step is to learn about percent yield. This value describes the efficiency of a chemical reaction. Percent yield is calculated by dividing the actual yield (what is practically obtained from a reaction) by the theoretical yield and then multiplying the result by 100 to get a percentage.

The formula is as follows:\[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\]
If we obtained \(3.0 \)mol of \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) (Freon) in the reaction mentioned, and we previously calculated the theoretical yield to be \(4.0 \)mol, the percent yield is \(\frac{3.0}{4.0} \times 100 = 75\)% which reflects that only 75% of the theoretical yield was achieved. This concept is a vital measure in gauging the practical limitations and efficiency of chemical processes in real-world applications, such as in pharmaceutical and chemical manufacturing industries. When answering such questions, accuracy in measurements and calculations can lead to a better understanding of the actual versus expected outcomes of reactions.

Limiting Reagent Concept

The limiting reagent concept is a cornerstone of reaction stoichiometry and is critical for predicting yields and understanding reaction dynamics. The limiting reagent is the reactant that is completely used up first during a chemical reaction, thus limiting the amount of product that can be formed.

In the provided exercise, \(\mathrm{CCl}_{4}\) is the limiting reagent since it is used in a defined amount and \(\mathrm{HF}\) is in excess. Therefore, \(\mathrm{CCl}_{4}\) dictates the maximum amount of product that can be formed. Subsequently, when analyzing a reaction, students should identify the limiting reagent as it determines the theoretical yield of the products. If a student can grasp this key concept, they'll be more adept at interpreting reaction parameters and predicting the outcome of reactions where not all reactants are in stoichiometric proportions.
It also must be noted that the absence of any \(\mathrm{CCl}_{4}\) at the reaction's conclusion, as stated in the exercise, is characteristic behavior for the limiting reagent. Recognizing the role of the limiting reagent enhances one's ability to comprehend and predict the flow of chemical reactions.