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March 29, 2017

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Asprin, C9H8O4, is prepared by reacting salicylic acid, C7H6O3, with acetic anhydride, C4H6O3, in the reaction
C7H6O3(s) + C4H6O3(l) -> C9H8O4(s) + C2H4O2(l)
A student is told to prepare 45.0 g of aspirin. She is also told to use a 55.0% excess of the acetic anhydride and to expect to get an 85.0% yield in the reaction. How many grams of each reactant should she use?


Can someone please give me a brief outline what I need to do to solve this?


You want 45 g aspirin. How many moles of asprin is that?

You want that number of moles/.85 of salicytic acid, and aspirin (numbermolesAspirin/.85) * 1.55 of the acetic anhydride. Think this out, repost if you have questions.

  • chemistry - ,

    C7H6O3 (s) + C4H6O3 (l) -> C9H8O4 (s) + C2H4O2 (l)
    molar masses:
    salicylic acid, C7H6O3 = 138.12 g/mol
    acetic anhydride, C4H6O3 = 102.09 g/mol
    Asprin, C9H8O4 = 180.157 g/mol

    For this problem calculate the moles of aspirin obtained from 45.0 g. This represents an 85.0% yield.

    % yield = (mass aspirin produced)/(theoretical mass aspirin)](100%)

    Using those numbers above, calculate the theoretical mass of aspirin needed. Once you have that number you can calculate the theoretical moles of aspirin. Use the chemical equation to determine the mass of salicylic acid needed to obtain the 45 grams.

    Moles aspirin = moles salicylic acid = moles acetic anhydride.

    Once you have the moles of acetic acid then you can determine the 55% excess by multiplying by 1.55(moles acetic anhydride).

  • chemistry - ,

    2 C7H6O3 + 1 C4H6O3 -> 2 C9H8O4 + H2O

    45.0 g / 0.85 = 52.9 g theoretical yield.

    52.9 g C9H8O4 x (1 mole / 180.2g) = 0.294 moles aspirin

    0.294 moles aspirin x (2 moles C7H6O3 / 2 moles aspirin) x (138.1 g C7H6O3 / mole C7H6O3) = 40.6 g salicylic acid.

    0.294 moles aspirin x (1 mole C4H6O3 / 2 moles aspirin) x (102.1 g C4H6O3 / mole C4H6O3) = 15.0 g acetic anhydride.

    but you need 55% excess C4H6O3... 1.55 x 15.0 g = 23.3 g acetic anhydride

  • chemistry - ,

    how'd you get 1.55g acetic anhydride?

  • chemistry - ,

    jbkjkj j

  • chemistry - ,

    I think the guy/girl was supposed to put .55 but accidently put a one insead of a zero and tried to figure it out like that. The actual answer is:
    0.55 x 15.0 g = 8.25 g acetic anhydride

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