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bruno102

  • one year ago

Check my answer? In the manufacturing process of sulfuric acid, sulfur dioxide is reacted with oxygen to produce sulfur trioxide. Using the equation, 2SO2 (g) + O2 arrow 2SO3 (g), if 128 g of sulfur dioxide is given the opportunity to react with an excess of oxygen, but only produces 144 g of sulfur trioxide, what is the percent yield of this reaction? 29.24% 58.48% 90.0% 100.0%

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  1. bruno102
    • one year ago
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    My answer is C

  2. pooja195
    • one year ago
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    Im not sure about this one :/ can you try looking at this? :/ http://openstudy.com/study#/updates/53a99966e4b0ffdda15ea9ee If we cant come up with anything ill tag someone else :)

  3. bruno102
    • one year ago
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    Hmm, I think I have it correct, but if you could tag someone, that would be helpful :)

  4. pooja195
    • one year ago
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    Ok :) @misssunshinexxoxo @Compassionate @chmvijay

  5. pooja195
    • one year ago
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    I think you have it right :/

  6. bruno102
    • one year ago
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    Thank you I hope so!

  7. pooja195
    • one year ago
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    :)

  8. bruno102
    • one year ago
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    @Ciarán95

  9. Elsa213
    • one year ago
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    What do you think the answer is?

  10. pooja195
    • one year ago
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    They put C

  11. pooja195
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @bruno102 My answer is C \(\color{blue}{\text{End of Quote}}\) .

  12. Elsa213
    • one year ago
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    e.e

  13. Elsa213
    • one year ago
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    I think you are correct, Bruno102.

  14. bruno102
    • one year ago
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    Thank you!

  15. iYuko
    • one year ago
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    Convert the 128g into moles

  16. bruno102
    • one year ago
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    @iYuko my answer isn't correct?

  17. iYuko
    • one year ago
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    You tell me.

  18. iYuko
    • one year ago
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    Multiply that by 2 then convert it to a mass

  19. bruno102
    • one year ago
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    Yes, I think so.

  20. bruno102
    • one year ago
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    @iYuko what should I do after I convert it to a mass?

  21. pooja195
    • one year ago
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    @iYuko

  22. taramgrant0543664
    • one year ago
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    What you do is you convert mass into moles of SO2, since the ratio of SO2 to SO3 is 2:2 you actually don't need to multiply by 2, so you have the moles of SO3 currently, so you are able to multiply that by the molar mass of SO3 to get your theoretical yield. So you can take your experimental yield and divide by your theoretical yield and multiply by 100 to get your percent

  23. taramgrant0543664
    • one year ago
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    And I got 90% when I did it

  24. bruno102
    • one year ago
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    Thank you so much @taramgrant0543664

  25. taramgrant0543664
    • one year ago
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    No problem!

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