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Ch.5 - Gases
Chapter 5, Problem 104b

Consider the reaction:
2 SO2(g) + O2(g) → 2 SO3(g)
b. If 187.2 mL of SO3 is collected (measured at 315 K and 50.0 mmHg), what is the percent yield for the reaction?

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1
Identify the balanced chemical equation: 2 SO_2(g) + O_2(g) \rightarrow 2 SO_3(g).
Use the ideal gas law, PV = nRT, to calculate the moles of SO_3 collected. Convert the pressure from mmHg to atm and the volume from mL to L.
Calculate the theoretical yield of SO_3 in moles using stoichiometry based on the initial amounts of reactants (not provided, but assume you have this information).
Convert the theoretical yield of SO_3 from moles to volume using the ideal gas law, assuming the same conditions of temperature and pressure.
Calculate the percent yield using the formula: \text{Percent Yield} = \left( \frac{\text{Actual Yield (in moles)}}{\text{Theoretical Yield (in moles)}} \right) \times 100.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced equation. It allows us to determine the theoretical yield of a product from given amounts of reactants. In this case, understanding the stoichiometric ratios of SO2 and O2 to SO3 is essential for calculating the expected yield.
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Percent Yield

Percent yield is a measure of the efficiency of a reaction, calculated by comparing the actual yield obtained from an experiment to the theoretical yield predicted by stoichiometry. It is expressed as a percentage and provides insight into how effectively a reaction proceeds under given conditions. The formula is: (actual yield / theoretical yield) × 100.
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Gas Laws

Gas laws describe the behavior of gases in relation to pressure, volume, and temperature. In this question, the ideal gas law (PV = nRT) can be used to convert the volume of SO3 collected into moles, which is necessary for calculating the percent yield. Understanding how to manipulate these variables is crucial for accurate calculations.
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