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Chapter 3: Problem 84

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\), the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\), but other products containing \(\mathrm{Cl}\), such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\), are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\). (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when \(125 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with \(255 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}\) and \(\mathrm{HCl}\). (b) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces \(206 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\).

Short Answer

Expert verified
Theoretical yield: 232.24 g; Percent yield: 88.70%.

Step by step solution

01

Write and Balance the Chemical Equation

The reaction between ethane (\(\mathrm{C}_2\mathrm{H}_6\)) and chlorine (\(\mathrm{Cl}_2\)) to form chloroethane (\(\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}\)) and hydrochloric acid (\(\mathrm{HCl}\)) is given by:\[\mathrm{C}_2\mathrm{H}_6 + \mathrm{Cl}_2 \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{Cl} + \mathrm{HCl}\] This equation is already balanced as written.
02

Calculate Molar Masses

Calculate the molar mass of each compound: - \(\mathrm{C}_2\mathrm{H}_6\): \(2(12.01) + 6(1.01) = 30.08\ \text{g/mol}\)- \(\mathrm{Cl}_2\): \(2(35.45) = 70.90\ \text{g/mol}\)- \(\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}\): \(2(12.01) + 5(1.01) + 35.45 = 64.51\ \text{g/mol}\)
03

Determine Moles of Reactants

Find the moles of \(\mathrm{C}_2\mathrm{H}_6\) and \(\mathrm{Cl}_2\).- Moles of \(\mathrm{C}_2\mathrm{H}_6\):\[\frac{125\ \text{g}}{30.08\ \text{g/mol}} \approx 4.15\ \text{mol}\]- Moles of \(\mathrm{Cl}_2\):\[\frac{255\ \text{g}}{70.90\ \text{g/mol}} \approx 3.60\ \text{mol}\]
04

Identify the Limiting Reactant

According to the balanced equation, 1 mole of \(\mathrm{C}_2\mathrm{H}_6\) reacts with 1 mole of \(\mathrm{Cl}_2\). Since there are 4.15 moles of \(\mathrm{C}_2\mathrm{H}_6\) and 3.60 moles of \(\mathrm{Cl}_2\), \(\mathrm{Cl}_2\) is the limiting reactant.
05

Calculate Theoretical Yield

Using the moles of the limiting reactant (\(\mathrm{Cl}_2\)), calculate the moles of \(\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}\) produced, which is also 3.60 moles. Convert to grams:\[3.60\ \text{mol} \times 64.51\ \text{g/mol} \approx 232.24\ \text{g}\] Thus, the theoretical yield is 232.24 g.
06

Calculate Percent Yield

The percent yield is given by \(\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\). Using the actual yield of 206 g:\[\frac{206\ \text{g}}{232.24\ \text{g}} \times 100 \approx 88.70\%%\] The percent yield is approximately 88.70%.

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Key Concepts

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

Limiting Reactant
In a chemical reaction, the limiting reactant is the substance that is totally consumed first, limiting the amount of product formed. It's like making sandwiches: if you have plenty of ingredients like fillings and spreads but run out of bread, you can't make more sandwiches. Here, chlorine (\(\mathrm{Cl}_2\)) is the limiting reactant. Even though there is more than enough ethane (\(\mathrm{C}_2\mathrm{H}_6\)), the reaction will stop when the chlorine is used up. This is because, according to the balanced chemical equation:\[- \mathrm{C}_2\mathrm{H}_6 + \mathrm{Cl}_2 \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{Cl} + \mathrm{HCl}\]One mole of ethane reacts with one mole of chlorine. We calculate and find:
  • 4.15 moles of \(\mathrm{C}_2\mathrm{H}_6\) are available.
  • 3.60 moles of \(\mathrm{Cl}_2\) are available.
Since chlorine runs out first, it limits the amount of chloroethane (\(\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}\)) produced.
Percent Yield
Percent yield is a measure of the efficiency of a reaction. It shows how much product was actually made compared to how much could have been theoretically produced (called the theoretical yield). In the world of chemistry, reactions often do not give 100% yield due to side reactions or incomplete reactions.For this reaction, the actual yield was found to be 206 g of \(\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}\). We calculate percent yield by using the formula:
  • Percent Yield = \[\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\]
By substituting the known values:\[\frac{206\ \text{g}}{232.24\ \text{g}} \times 100 \approx 88.70\%\]So, the percent yield is about 88.70%, which indicates a fairly efficient reaction, meaning most of the reactants were successfully converted into the desired product.
Theoretical Yield
The theoretical yield of a reaction is the amount of product that would be formed if everything went perfectly 鈥� meaning complete reaction of the limiting reactant without any losses or side reactions. It is calculated based on stoichiometry, assuming all the limiting reactant converts to the desired product.To find the theoretical yield in this particular reaction between ethane and chlorine:
  • We start by identifying that \(\mathrm{Cl}_2\) is the limiting reactant, with 3.60 moles available.
  • According to the balanced equation, 1 mole of \(\mathrm{Cl}_2\) produces 1 mole of \(\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}\).
  • Therefore, 3.60 moles of \(\mathrm{Cl}_2\) can produce 3.60 moles of \(\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}\).
  • Convert moles of \(\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}\) to grams: \[3.60\ \text{mol} \times 64.51\ \text{g/mol} = 232.24\ \text{g}\]
Thus, the theoretical yield is 232.24 g of chloroethane, assuming ideal conditions with no loss of reactants.

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Most popular questions from this chapter

A sample of the male sex hormone testosterone, \(\mathrm{C}_{19} \mathrm{H}_{28} \mathrm{O}_{2}\), contains \(3.88 \times 10^{21}\) hydrogen atoms. (a) How many atoms of carbon does it contain? (b) How many molecules of testosterone does it contain? (c) How many moles of testosterone does it contain? (d) What is the mass of this sample in grams?

Write "true" or "false" for each statement. (a) We balance chemical equations as we do because energy must be conserved. (b) If the reaction \(2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g)\) goes to completion and all \(\mathrm{O}_{3}\) is converted to \(\mathrm{O}_{2}\), then the mass of \(\mathrm{O}_{3}\) at the beginning of the reaction must be the same as the mass of \(\mathrm{O}_{2}\) at the end of the reaction. (c) You can balance the "water-splitting" reaction \(\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) by writing it this way: \(\mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\).

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to NO: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(2.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.50 \mathrm{~g}\) of \(\mathrm{O}_{2} .\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and \(\mathrm{H}_{2} \mathrm{O}\) form? \((\mathbf{c})\) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Write a balanced chemical equation for the reaction that occurs when (a) \(\mathrm{Mg}(s)\) reacts with \(\mathrm{Cl}_{2}(g) ;\) (b) barium carbonate decomposes into barium oxide and carbon dioxide gas when heated; \((\mathbf{c})\) the hydrocarbon styrene, \(\mathrm{C}_{8} \mathrm{H}_{8}(l),\) is combusted in air; \((\mathbf{d})\) dimethylether, \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(g),\) is combusted in air.

(a) When a compound containing C, H, and O is completely combusted in air, what reactant besides the hydrocarbon is involved in the reaction? (b) What products form in this reaction? (c) What is the sum of the coefficients in the balanced chemical equation for the combustion of one mole of acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l),\) in air?
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