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The text explains that one reason why the actual yicld for a reaction may be less than the theoretical yicld is side reactions. Suggest some other reasons why the percent yield for a reaction might not be \(100 \%\).

Short Answer

Expert verified
Some reasons for a percent yield less than \(100\%\) include: 1) Incomplete reaction, possibly due to insufficient time or equilibrium; 2) Loss of product during extraction or purification; 3) Impure reactants causing side reactions or non-participation; 4) Experimental errors, such as inaccurate measurements or incorrect reaction conditions; and 5) Limiting reactants, which restrict the formation of the desired product.

Step by step solution

01

Reason 1: Incomplete Reaction

One reason the percent yield might not be 100% is that the reaction might not go to completion. This could be due to insufficient time for the reaction to occur, or because the reaction reaches a state of equilibrium where it stops reacting.
02

Reason 2: Loss of Product during Extraction or Purification

Another reason is the loss of product during the process of extraction or purification. This can happen during separation, filtration, or other steps involved in isolating the desired product. Some product might be lost, which would lower the yield.
03

Reason 3: Impure Reactants

If the reactants used for the reaction are impure, it could cause a lower percent yield. The impurities present may not participate in the reaction or may cause side reactions that consume some of the desired product.
04

Reason 4: Experimental Errors

There could also be experimental errors when performing the reaction. These errors might include inaccurate measurements of reactants, incorrect reaction conditions (e.g., temperature, pressure), or other mistakes during the setup and execution of the reaction.
05

Reason 5: Limiting Reactants

If one of the reactants is present in a smaller amount than required for a complete reaction, it is called the limiting reactant. The limiting reactant can restrict the formation of the desired product, resulting in a lower percent yield. These are some fitting reasons why the percent yield for a chemical reaction might not be 100%. In reality, the actual percent yield is typically lower than the theoretical yield due to any combination of these factors.

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Key Concepts

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

Side Reactions
Side reactions are unintended chemical reactions that occur alongside the main reaction. These reactions can sometimes consume the reactants that were meant to produce the desired product. This can decrease the amount of the main product and therefore impact the percent yield.
  • Side reactions can create byproducts that may not be useful or wanted.
  • These reactions can often compete with the primary reaction for reactants.
It's essential to be aware of potential side reactions when designing a reaction in the lab. Chemists often use conditions that favor the main reaction while minimizing the side reactions to improve percent yield.
Incomplete Reaction
An incomplete reaction can occur when the reaction does not go all the way to forming the expected amount of product. This is typically due to the reaction reaching equilibrium before it is complete, or because it doesn't have enough time to go to completion.
  • In some cases, not all reactants have had a chance to react. This means that some reactants are left unconverted.
  • Different factors like reaction time, temperature, and pressure can influence the completion of a reaction.
Understanding the kinetics and dynamics of a reaction can help in adjusting these conditions to maximize product yield.
Impure Reactants
Using impure reactants in a chemical reaction can have a significant impact on the percent yield. Impurities in reactants do not participate efficiently in the reaction, which affects the overall amount of product formed.
  • Impure reactants can lead to side reactions that consume some reactants or even the product.
  • Purify reactants beforehand to ensure accuracy in the reaction outputs.
Determining the purity of reactants before proceeding with the reaction is crucial for obtaining a more accurate and reliable yield.
Experimental Errors
Experimental errors are also a common cause of reduced percent yield. These errors might include mistakes in measuring reactants, setting reaction conditions, or handling products.
  • Errors in measurement can lead to incorrect proportions of reactants.
  • Inaccurate temperatures or pressures can also influence the reaction's success.
Taking precise measurements and carefully following protocols can help minimize experimental errors and maximize percent yield.

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Most popular questions from this chapter

Explain how one determines which reactant in a process is the limiting reactant. Does this depend only on the masses of the reactant present? Give an example of how to determine the limiting reactant by using a Before-Change- After (BCA) table with a balanced chemical cquation and reactant starting amounts.

Natural waters often contain relatively high levels of calcium ion, \(\mathrm{Ca}^{2+},\) and hydrogen carbonate ion (bicarbonate), \(\mathrm{HCO}_{3}^{-}\), from the leaching of minerals into the water. When such water is used commercially or in the home, heating of the water leads to the formation of solid calcium carbonate, \(\mathrm{CaCO}_{3}\), which forms a deposit ("scale") on the interior of boilers, pipes, and other plumbing fixtures. $$ \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(a q) \rightarrow \mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If a sample of well water contains \(2.0 \times 10^{-3} \mathrm{mg}\) of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) per milliliter, what mass of \(\mathrm{CaCO}_{3}\) scale would \(1.0 \mathrm{~mL}\) of this water be capable of depositing?

One process for the commercial production of baking soda (sodium hydrogen carbonate) involves the following reaction, in which the carbon dioxide is used in its solid form ("dry ice" enough for the sodium hydrogen carbonate to precipitate: $$ \mathrm{NaCl}(a q)+\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(s) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{NaHCO}_{3}(s) $$ Because they are relatively cheap, sodium chloride and water are typically present in excess. What is the expected yield of \(\mathrm{NaHCO}_{3}\) when one performs such a synthesis using \(10.0 \mathrm{~g}\) of ammonia and \(15.0 \mathrm{~g}\) of dry ice, with an excess of \(\mathrm{NaCl}\) and water?

For cach of the following balanced reactions, calculate how many moles of each product would be produced by complete conversion of 0.50 mole of the reactant indicated in boldface. Indicate clearly the mole ratio used for the conversion. a. \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) b. \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) c. \(2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \rightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2}(g)\) d. \(\mathbf{C}_{3} \mathbf{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

A \(0.4230-g\) sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to \(0.2339 \mathrm{~g}\) of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

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