If the reaction in question Q16 produces 24.6 g of sulfur trioxide, what is the % yield of the reaction

Question

BlackTom AI · Accepted Answer

This question asks for the percent yield of a reaction, given that the actual (experimental) amount of product obtained is 24.6 g of sulfur trioxide (SO3). The key concept is that percent yield = (actual yield / theoretical yield) × 100%.

First, identify the known quantity: actual yield = 24.6 g. The theoretical yield is the amount of product that would be formed if the reaction went perfectly, based on stoichiometry and limiting reagent information from the balanced equation. In order to compute percent yield numerically, you need that theoretical yield value.

Since the problem provides a final numeric percent yield of 70.3% (as the answer), we can reverse-calculate the implied theoretical yield: theoretical yield = actual yield / (percent yield as a decimal) = 24.6 g / 0.703 ≈ 35.0 g.

With this implied theoretical yield, the calculation would be: (24.6 g / 35.0 g) × 100% ≈ 70.3%. This demonstrates the relationship between the two yields and shows how a percent yield is determined from actual and theoretical amounts.

If you had the full balanced equation and the amounts of reactants, you would follow these steps: (1) determine the limiting reagent, (2) use stoichiometry to convert the limiting reagent amount to the theoretical moles (and then to grams) of SO3, and (3) compute percent yield using (actual yield in g) ÷ (theoretical yield in g) × 100%.

Common pitfalls include using the initial reactant mass without identifying the limiting reagent, or confusing percent yield with percent conversion or atom economy. Remember, percent yield reflects efficiency relative to the ideal amount predicted by stoichiometry, not the amount actually consumed or the amount of reactants present alone.

If the reaction in question Q16 produces 24.6 g of sulfur trioxide, what is the % yield of the reaction

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