Answer :
The percent yield of silver in the reaction is 85.0%.
What is the percentage of silver obtained in this reaction?
In the given reaction between copper metal (Cu) and silver nitrate (AgNO₃), 12.7 grams of copper produced 38.1 grams of silver. To calculate the percent yield of silver, we need to compare the actual yield (38.1 g) with the theoretical yield, which is the maximum amount of silver that could be produced based on the stoichiometry of the reaction.
The balanced chemical equation for the reaction is:
2AgNO₃ + Cu → Cu(NO₃)₂ + 2Ag
From the equation, we can see that for every 1 mole of copper (Cu), 2 moles of silver (Ag) are produced. To determine the theoretical yield of silver, we need to convert the mass of copper (12.7 g) to moles using its molar mass (63.55 g/mol), and then use the stoichiometry to calculate the corresponding mass of silver.
12.7 g Cu × (1 mol Cu / 63.55 g Cu) × (2 mol Ag / 1 mol Cu) × (107.87 g Ag / 1 mol Ag) = 38.1 g Ag
Therefore, the theoretical yield of silver is 38.1 grams.
To calculate the percent yield, we divide the actual yield (38.1 g) by the theoretical yield (38.1 g) and multiply by 100:
Percent yield = (Actual yield / Theoretical yield) × 100
= (38.1 g / 38.1 g) × 100
= 100%
Thus, the percent yield of silver in this reaction is 100%.
The percent yield of a chemical reaction is a measure of the efficiency of the reaction in producing the desired product. It indicates the proportion of the theoretical yield that was actually obtained in the experiment. A percent yield greater than 100% suggests that more product was obtained than predicted, which could be due to experimental errors, impurities, or side reactions. In this case, the percent yield of silver is 100%, indicating that the reaction proceeded with high efficiency and all the available copper was converted into silver.
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