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anonymous

  • one year ago

When 24.9 grams of ethane (C2H6) react with excess oxygen to produce carbon dioxide, the percent yield is 56.8%. What was the actual yield of the reaction? 2C2H6 + 7O2 yields 4CO2 + 6H2O 20.7 g 41.5 g 62.2 g 64.3 g

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  1. aaronq
    • one year ago
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    First we need to find the theoretical yield, then multiply it by the yield percentage. So, you need to convert the mass to moles. Use: \(\sf moles=\dfrac{mass}{Molar~mass}\) Next we need a ratio of the moles of what you're using and what you're producing, and their respective stoichiometric coefficients from the balanced reaction (the numbers preceding the chemical formulas): \(\sf\dfrac{moles~of~ethane}{ethane's~coefficient}=\dfrac{moles~of~CO_2}{CO_2's~coefficient}\) Plug in moles of ethane and the coefficients and solve for moles of carbon dioxide. This is the theoretical yield, the MOST you could produce with what you have. Now multiply by 0.568. Done

  2. anonymous
    • one year ago
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    can you just solve it please @aaronq

  3. aaronq
    • one year ago
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    Nah, this is a helping site, you have to do it yourself.

  4. anonymous
    • one year ago
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    well most of the people on her just leave the answer

  5. aaronq
    • one year ago
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    if this was hard I would show you by doing it, but it's really easy.

  6. anonymous
    • one year ago
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    okay tell me all the caluculations and ill calutate thenm to find the answer @aaronq

  7. aaronq
    • one year ago
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    I just explained in detail how to do it above, I even gave you the formulas

  8. anonymous
    • one year ago
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    @aaronq well do me this favor and solve have 2 minutes left on this test

  9. aaronq
    • one year ago
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    you're not putting enough effort forth, so i can't help you

  10. anonymous
    • one year ago
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    @aaronq eat retricefeather

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