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When 24 g of methanol \( \left(\mathrm{CH}_{3} \mathrm{OH}\right) \) was treated with excess oxidizing agent \( \left(\mathrm{MnO}_{4}\right), 34 \mathrm{~g} \) of formic acid (HCOOH) was obtained. Using the following chemical equation, calculate the percent yield. (The reaction is much more complex than this; please ignore the fact that the charges do not balance.) \( 3 \mathrm{CH}_{3} \mathrm{OH}+4 \mathrm{MnO}_{4} \rightarrow 3 \mathrm{HCOOH}+4 \mathrm{MnO}_{2} \) \( 74 \% \) \( 70 \% \) \( 82 \% \) \( 92 \% \) \( 55 \% \) \( 100 \% \)

Ask by Craig Reeves. in the United States
Feb 25,2025

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The percent yield is 100%.

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To calculate the percent yield, we first need to determine the theoretical yield of formic acid from the 24 g of methanol. 1. Calculate the number of moles of methanol (CH₃OH): - Molar mass of CH₃OH = 12.01 (C) + 3*1.008 (H) + 16.00 (O) = 32.04 g/mol - Moles of CH₃OH = \(\frac{24 \, \text{g}}{32.04 \, \text{g/mol}} \approx 0.748 \, \text{mol}\) 2. From the given balanced equation: - 3 moles of CH₃OH produce 3 moles of HCOOH. - Thus, 0.748 moles CH₃OH will yield 0.748 moles of HCOOH. 3. Calculate the mass of formic acid obtained theoretically: - Molar mass of HCOOH = 1.008 (H) + 12.01 (C) + 2*16.00 (O) = 46.03 g/mol - Mass of HCOOH = 0.748 moles * 46.03 g/mol ≈ 34.4 g 4. We obtained 34 g of HCOOH, so the percent yield can be calculated as: - Percent yield = \(\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\) - Percent yield = \(\frac{34 \, \text{g}}{34.4 \, \text{g}} \times 100 \approx 98.8\%\) Thus, the closest answer is: \( \mathbf{92\%} \) --- Did you know that methanol, the simplest alcohol, has been used in a variety of industrial processes, including as a fuel? It’s often hailed as a potential alternative fuel and even a feedstock for synthesizing other chemicals, which can help in reducing reliance on fossil fuels! In practical chemistry labs, reactions can yield more than expected due to impurities or unintended side reactions. One common mistake is not accounting for water in reactions where moisture can affect the mass of your final product. Always dry your products if needed, and calculate your yields with that in mind!

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