/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 142 In the reaction of \(2.00 \mathr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the reaction of \(2.00 \mathrm{mol} \mathrm{CCl}_{4}\) with an excess of \(\mathrm{HF}\), \(1.70 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) is obtained. $$ \mathrm{CCl}_{4}+2 \mathrm{HF} \longrightarrow \mathrm{CCl}_{2} \mathrm{F}_{2}+2 \mathrm{HCl} $$ (a) The theoretical yield is \(1.70 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) (b) The theoretical yield is \(1.00 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) (c) The theoretical yield depends on how large an excess of HF is used. (d) The percent yield is \(85 \%\)

Short Answer

Expert verified
(a) The theoretical yield is 2.00 mol \( \mathrm{CCl}_{2}\mathrm{F}_{2} \) (d) The percent yield is \( 85 \% \)

Step by step solution

01

Theoretical Yield Calculation

The theoretical yield is determined from the stochiometric coefficients of reactants and products in the balanced chemical reaction. In this reaction: \( \mathrm{CCl}_4 + 2\mathrm{HF} \rightarrow \mathrm{CCl}_2\mathrm{F}_2 + 2\mathrm{HCl} \) , for every one mole of \( \mathrm{CCl}_4 \) reacted, one mole of \( \mathrm{CCl}_2\mathrm{F}_2 \) is produced. Hence, the theoretical yield of \( \mathrm{CCl}_2\mathrm{F}_2 \) when 2.00 mol of \( \mathrm{CCl}_4 \) are reacted is 2.00 mol.
02

Percent yield Calculation

Percent yield is the ratio of the actual yield to the theoretical yield expressed as a percentage. It is calculated using the formula: \( \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 \) Here, the actual yield is 1.70 mol and the theoretical yield is 2.00 mol. Thus, the percent yield is \( \frac{1.70\, \text{mol}}{2.00\, \text{mol}} \times 100 = 85 \% \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Percent Yield
Percent yield is an important concept in chemistry as it tells us how efficient a chemical reaction has been. The percent yield provides insight into how much product was actually obtained from a reaction compared to the maximum expected amount, known as the theoretical yield. To calculate percent yield, we use the formula:
  • Percent Yield = \( \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \)
In the given problem, the actual yield of \( \text{CCl}_2\text{F}_2 \) is stated to be 1.70 mol, whereas the theoretically possible amount is 2.00 mol. Therefore, calculating the percent yield shows us that the reaction had a yield of 85%.
This means that only 85% of the maximum possible product was produced, implying some inefficiencies or experimental conditions that prevented a full conversion of reactants to products.
Chemical Reaction
Understanding chemical reactions entails knowing how and why reactants convert into products. In chemical equations, reactants are transformed into products when their molecules interact. For instance, the reaction between carbon tetrachloride \((\mathrm{CCl}_4)\) and hydrogen fluoride \((\mathrm{HF})\) is described by the equation:
  • \( \mathrm{CCl}_4 + 2 \mathrm{HF} \rightarrow \mathrm{CCl}_2\mathrm{F}_2 + 2\mathrm{HCl} \)
This equation shows that one molecule of \(\mathrm{CCl}_4\) reacts with two molecules of \(\mathrm{HF}\) to produce one molecule of \(\mathrm{CCl}_2\mathrm{F}_2\) and two molecules of \(\mathrm{HCl}\). The coefficients before each compound indicate the proportion of molecules involved in the reaction.
Chemical reactions require specific conditions, such as temperature, pressure, or catalysts, to proceed efficiently. Reactants must be present in adequate amounts, as shown by the reaction's stoichiometry, to maximize the production of products.
Stoichiometry
Stoichiometry plays a key role in understanding chemical reactions, as it relates to the quantitative relationships of reactants and products. It allows chemists to predict how much product can be formed from given amounts of reactants. This is where the concept of theoretical yield comes in.
In the reaction \(\mathrm{CCl}_4 + 2\mathrm{HF} \rightarrow \mathrm{CCl}_2\mathrm{F}_2 + 2\mathrm{HCl}\), the stoichiometric coefficients tell us that one mole of \(\mathrm{CCl}_4\) reacts with two moles of \(\mathrm{HF}\) to produce one mole of \(\mathrm{CCl}_2\mathrm{F}_2\).
To calculate the theoretical yield, you'd use these stoichiometric ratios. With 2.00 moles of \(\mathrm{CCl}_4\) and assuming an excess of \(\mathrm{HF}\), it tells us the maximum amount of \(\mathrm{CCl}_2\mathrm{F}_2\) that can be formed is 2.00 moles.
  • This calculation helps predict how much product should form, assuming perfect conditions.
  • Any deviation from the theoretical yield hints at reaction inefficiencies or measurement inaccuracies, making stoichiometry crucial for accurate chemical predictions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How many grams of sodium must react with \(155 \mathrm{mL}\) \(\mathrm{H}_{2} \mathrm{O}\) to produce a solution that is \(0.175 \mathrm{M} \mathrm{NaOH} ?\) (Assume a final solution volume of \(155 \mathrm{mL}\) ) $$ 2 \mathrm{Na}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

Nitric acid, \(\mathrm{HNO}_{3}\), can be manufactured from ammonia, \(\mathrm{NH}_{3}\), by using the three reactions shown below. $$\begin{aligned} &\text { Step 1: 4 NH }_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\\\ &\text { Step 2: } 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\\\ &\text { Step 3: } 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{aligned}$$ What is the maximum number of moles of \(\mathrm{HNO}_{3}\) that can be obtained from 4.00 moles of \(\mathrm{NH}_{3}\) ? (Assume that the NO produced in step 3 is not What is the maximum number of moles of \(\mathrm{HNO}_{3}\) that can be obtained from 4.00 moles of \(\mathrm{NH}_{3}\) ? (Assume that the NO produced in step 3 is not recycled back into step 2.) (a) 1.33 mol; (b) 2.00 mol; (c) 2.67 mol; (d) 4.00 mol; (e) 6.00 mol.

Write chemical equations to represent the following reactions. (a) Calcium phosphate is heated with silicon dioxide and carbon, producing calcium silicate \(\left(\mathrm{CaSiO}_{3}\right)\) phosphorus ( \(\mathrm{P}_{4}\) ), and carbon monoxide. The phosphorus and chlorine react to form phosphorus trichloride, and the phosphorus trichloride and water react to form phosphorous acid. (b) Copper metal reacts with gaseous oxygen, carbon dioxide, and water to form green basic copper carbonate, \(\mathrm{Cu}_{2}(\mathrm{OH})_{2} \mathrm{CO}_{3}\) (a reaction responsible for the formation of the green patina, or coating, often seen on outdoor bronze statues). (c) White phosphorus and oxygen gas react to form tetraphosphorus decoxide. The tetraphosphorus decoxide reacts with water to form an aqueous solution of phosphoric acid. (d) Calcium dihydrogen phosphate reacts with sodium hydrogen carbonate (bicarbonate), producing calcium phosphate, sodium hydrogen phosphate, carbon dioxide, and water (the principal reaction occurring when ordinary baking powder is added to cakes, bread, and biscuits).

Under appropriate conditions, copper sulfate, potassium chromate, and water react to form a product containing \(\mathrm{Cu}^{2+},\) \(\mathrm{CrO}_{4}{^2}{^-},\) and \(\mathrm{OH}^{-}\) ions. Analysis of the compound yields \(48.7 \% \mathrm{Cu}^{2+}, 35.6 \% \mathrm{CrO}_{4}{^2}{-},\) and \(15.7 \% \mathrm{OH}^{-}\). (a) Determine the empirical formula of the compound. (b) Write a plausible equation for the reaction.

What are the molarities of the following solutes? (a) aspartic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{5} \mathrm{NO}_{4}\right)\) if \(0.405 \mathrm{g}\) is dissolved in enough water to make \(100.0 \mathrm{mL}\) of solution (b) acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O},(d=0.790 \mathrm{g} / \mathrm{mL})\) if \(35.0 \mathrm{mL}\) is dissolved in enough water to make \(425 \mathrm{mL}\) of solution (c) diethyl ether, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O},\) if \(8.8 \mathrm{mg}\) is dissolved in enough water to make 3.00 L of solution

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.