Problem 84 - Chemistry The Central Science In Si Units

Question

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 \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\).

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: Calculate Molar Masses

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

Step 3: 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 \(\mathrm{Cl}_2\):\[\frac{255\ \text{g}}{70.90\ \text{g/mol}} \approx 3.60\ \text{mol}\]

Step 4: 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.

Step 5: 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.

Step 6: 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%.

Key Concepts

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 — 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.