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Two successive reactions, \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{C},\) have yields of \(73 \%\) and \(68 \%,\) respectively. What is the overall percent yield for conversion of A to C?

Short Answer

Expert verified
The overall percent yield for conversion of A to C is 49.64%.

Step by step solution

01

- Identify the Given Data

The yields for two successive reactions are given. The first reaction, \(\text{A} \to \text{B}\), has a yield of 73%, and the second reaction, \(\text{B} \to \text{C}\), has a yield of 68%.
02

- Convert Percent Yields to Decimal Form

Convert each percent yield to decimal form by dividing by 100. \( 73\text{\text{%}} = 0.73 \) and \( 68\text{\text{%}} = 0.68 \).
03

- Multiply the Decimal Yields

Multiply the decimal forms of the yields to find the overall yield: \( 0.73 \times 0.68 \).
04

- Calculate the Overall Yield

Perform the multiplication: \( 0.73 \times 0.68 = 0.4964 \).
05

- Convert the Result Back to Percent Form

Convert the decimal result back to a percent by multiplying by 100: \( 0.4964 \times 100 = 49.64\text{\text{%}} \).

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Key Concepts

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

headline of the respective core concept
Successive reactions are reactions that happen one after another. In this context, we have two successive reactions: \(\text{A} \to \text{B} \) and \( \text{B} \to \text{C} \). Each step in a successive reaction can have a different percent yield. This determines how much of the initial substance eventually converts into the final product. It's essential to understand that the yields of successive reactions multiply to give the overall yield from the initial reactant to the final product.

To break this down:
  • The yield of the first reaction, \(\text{A} \to \text{B}\), is 73%.
  • The yield of the second reaction, \(\text{B} \to \text{C}\), is 68%.
To find the overall yield of \( \text{A} \to \text{C} \), we convert each percentage to a decimal by dividing by 100, then multiply them together:

\( 0.73 \times 0.68 = 0.4964 \).

This 0.4964 in decimal form translates back to 49.64% as the overall percent yield. So, overall percent yield for successive reactions is the product of individual yields.
headline of the respective core concept
Percent yield is a measure of how efficient a reaction is. It compares the amount of product actually obtained to the amount that theoretically could be produced. The formula to calculate percent yield is:

\[ \text{Percent Yield} = \left( \frac{ \text{Actual Yield} }{ \text{Theoretical Yield} } \right) \times 100 \]

Here's why percent yield matters:
  • It helps chemists understand the efficiency of a reaction.
  • It identifies how much product loss occurs during the reaction.
  • High percent yields indicate efficient reactions, minimal waste, and cost-effectiveness.
In our problem, the yields were given directly, but in practice, you determine them by comparing the actual product obtained (e.g., via measurement) to a calculated theoretical yield.
headline of the respective core concept
Reaction yields refer to how much product is made in a chemical reaction. This varies due to factors like reaction conditions, completeness, and side reactions. Calculating reaction yields can help optimize and scale-up processes. Here are the key points:
  • Theoretical Yield: The maximum possible amount of product, based on stoichiometry.
  • Actual Yield: The actual amount of product obtained, which is often less due to inefficiencies.
  • Percent Yield: A ratio of actual yield to theoretical yield, expressed as a percentage.
Knowing reaction yields is critical in fields like pharmaceuticals, where precise amounts are crucial. They also highlight which steps in a process might need improvement to minimize losses and maximize efficiencies.

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Most popular questions from this chapter

A 0.370 -mol sample of a metal oxide \(\left(\mathrm{M}_{2} \mathrm{O}_{3}\right)\) weighs \(55.4 \mathrm{~g}\). (a) How many moles of \(\mathrm{O}\) are in the sample? (b) How many grams of \(\mathrm{M}\) are in the sample? (c) What element is represented by the symbol M?

What are the empirical formula and empirical formula mass for each of the following compounds? (a) \(\mathrm{C}_{4} \mathrm{H}_{8}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (c) \(\mathrm{P}_{4} \mathrm{O}_{10}\) (d) \(\mathrm{Ga}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (e) \(\mathrm{Al}_{2} \mathrm{Br}_{6}\)

What is the difference between an empirical formula and a molecular formula? Can they ever be the same?

Narceine is a narcotic in opium that crystallizes from solution as a hydrate that contains 10.8 mass \(\%\) water and has a molar mass of \(499.52 \mathrm{~g} / \mathrm{mol} .\) Determine \(x\) in narceine \(x \mathrm{H}_{2} \mathrm{O}\).

Various nitrogen oxides, as well as sulfur oxides, contribute to acidic rainfall through complex reaction sequences. Nitrogen and oxygen combine during the high-temperature combustion of fuels in air to form nitrogen monoxide gas, which reacts with more oxygen to form nitrogen dioxide gas. In contact with water vapor, nitrogen dioxide forms aqueous nitric acid and more nitrogen monoxide. (a) Write balanced equations for these reactions. (b) Use the equations to write one overall balanced equation that does not include nitrogen monoxide and nitrogen dioxide. (c) How many metric tons (t) of nitric acid form when \(1350 \mathrm{t}\) of atmospheric nitrogen is consumed \((1 \mathrm{t}=1000 \mathrm{~kg}) ?\)

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