Give two reasons why the actual yield from chemical reactions is less than 100\(\% .\)

Short Answer

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The actual yield from chemical reactions is often less than 100% due to two primary reasons: (1) Not all reactions go to completion, especially those that are reversible. (2) Loss of product can occur during the isolation and purification steps, where some of the product may be lost due to physical processes like filtration, evaporation, and distillation.

Step by step solution

01

Understanding the concept of 'yield'

In chemistry, 'yield' refers to the amount of product that is produced in a chemical reaction. Theoretical yield refers to the maximum amount of product that can be produced from a given amount of reactants, as predicted by the stoichiometric ratio of the reaction. Actual yield, on the other hand, refers to the amount of product that is actually formed when the reaction is carried out. In practice, the actual yield is often less than the theoretical yield.
02

Reason 1: Not All Reactions Go to Completion

One reason why the actual yield is less than the theoretical yield is that not all reactions go to completion. Some reactions, especially reactions that are reversible, can stop before all the reactants have been converted into products.
03

Reason 2: Loss of Product

Another reason for the difference between actual yield and theoretical yield is the loss of product. This can occur during the isolation and purification steps of the process, where some of the product may be lost due to physical processes such as filtration, evaporation, and distillation.

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Key Concepts

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

Theoretical Yield
When embarking on a chemical reaction, scientists and students alike aim to predict the maximum amount of product that can be generated from a given quantity of reactants. This is known as the theoretical yield. It is calculated based on the stoichiometry of a balanced chemical equation and assumes perfect conditions where everything reacts exactly as expected.

To put it simply, if we were baking a cake and had a precise recipe, the theoretical yield would represent the ideal number of cakes we could bake from our ingredients, without any mixture left over or any cake wrecked in the process.

In reality, achieving the theoretical yield is like baking the perfect cake every single time, without fail—it's an ideal situation that often isn't met due to various practical factors. Understanding theoretical yield is crucial for chemists, as it sets the benchmark against which the actual efficiency of a reaction can be measured. It’s an aspirational figure that reveals how efficient a reaction could be under optimal conditions.
Chemical Reaction Completion
The journey of reactants to become products isn't always a smooth one. One major reason why the actual yield in chemical reactions falls short of the theoretical yield is the issue of reaction completion. Not all reactions proceed to the point where all reactants are fully converted into products.

This can be particularly true for reversible reactions, where the products can react to form the reactants again. Such reactions may reach a state of equilibrium, where the rate of the forward reaction equals the rate of the reverse reaction, thus halting any further net progress.

Imagine our cake analogy again; some of the reactants—the ingredients—might not fully mix or react as expected, leaving us with a less than perfect cake, or in chemistry terms, a lower actual yield of product. In practice, chemists work to optimize reaction conditions, such as temperature, pressure, and catalyst presence, to drive the reaction as close to completion as possible and inch closer to that elusive 100% yield.
Product Loss during Isolation and Purification
Getting the desired product from a chemical reaction doesn't end with the completion of the reaction itself. There's another hurdle to cross: collecting the product, which often involves various steps such as filtration, distillation, or crystallization.

During these isolation and purification processes, not all of the product makes it out of the chemical flask and into the final container. Some of it can stick to the filters, remain undissolved in the mother liquor, get lost as vapors during distillation, or simply degrade over time.

In our cake-baking scenario, it's akin to part of the cake sticking to the baking pan or some of it crumbling away when removing it from the pan. Such losses may seem minor in a single instance, but in the scale of industrial processes, they can represent a significant amount of product—and potential profit—gone to waste. Chemists therefore are not only concerned with the chemical aspects of the reaction, but also with developing efficient techniques for product recovery to minimize such losses and maximize the actual yield.

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Most popular questions from this chapter

Iron and CO are made by heating 4.56 \(\mathrm{kg}\) of iron ore, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) and carbon. The yield of iron is 88\(\% .\) How many kilograms of iron are made?

How many liters \(\mathrm{N}_{2},\) density \(0.92 \mathrm{g} / \mathrm{L},\) can be made by the decomposition of 2.05 \(\mathrm{g} \mathrm{NaN}_{3} ?\) $$2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)$$

In the reaction shown below, 64 \(\mathrm{g} \mathrm{CaC}_{2}\) is reacted with 64 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s)$$ \begin{equation}\begin{array}{l}{\text { a. Which is the excess reactant, and which is }} \\ {\text { the limiting reactant? }} \\ {\text { b. What is the theoretical yield of } C_{2} \mathrm{H}_{2} ?} \\ {\text { c. What is the theoretical yield of } \mathrm{Ca}(\mathrm{OH})_{2} ?}\end{array}\end{equation}

Why is it necessary to use mole ratios in solving stoichiometry problems?

The following reaction can be used to remove \(\mathrm{CO}_{2}\) breathed out by astronauts in a spacecraft. $$2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)$$ \begin{equation}\begin{array}{l}{\text { a. How many grams of carbon dioxide can }} \\ {\text { be removed by } 5.5 \text { mol LiOH? }} \\ {\text { b. How many milliliters } \mathrm{H}_{2} \mathrm{O} \text { (density }=} \\\ {0.997 \mathrm{g} / \mathrm{mL} \text { ) could form from } 25.7 \mathrm{g} \text { LiOH? }} \\ {\text { c. How many molecules } \mathrm{H}_{2} \mathrm{O} \text { could be }} \\ {\text { made when } 3.28 \mathrm{g} \mathrm{CO}_{2} \text { react? }}\end{array}\end{equation}

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