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    For the balanced equation shown below, if the reaction of 0.112 grams of H2 produces 0.745 grams of H2O, what is the percent yield?

    Fe3O4+4H2=>3Fe+4H2O

    Here's what I did:

    mol of H2: 0.112/(2 x 4) = 0.014

    mol of H2O: 0.74/(18 x 4) = 0.0103

    0.0103/0.014 x 100 = 73.6 %


    This is what they did (, I know their working out is a little different from mine, but I'm using the one my teacher showed me). Can someone please tell me where I went wrong?

    H = 1
    O = 16

    1 x 2 = 2 grams/mole

    1 x 2 = 2
    16 x 1 = 16

    2 + 16 = 18 grams/mole

    2 x 4 = 8 18 x 4 = 72

    72/8 = 9

    9 x 0.112 = 1

    percentage yield = (0.745/1) x 100 = 74.5 %

    percentage yield = 74.5%
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    (Original post by celina10)
    For the balanced equation shown below, if the reaction of 0.112 grams of H2 produces 0.745 grams of H2O, what is the percent yield?

    Fe3O4+4H2=>3Fe+4H2O

    Here's what I did:

    mol of H2: 0.112/(2 x 4) = 0/014

    mol of H2O: 0.74/(18 x 4) = 0/0103

    0.0103/0.014 x 100 = 73.6 %


    This is what they did (, I know their working out is a little different from mine, but I'm using the one my teacher showed me). Can someone please tell me where I went wrong?

    H = 1
    O = 16

    1 x 2 = 2 grams/mole

    1 x 2 = 2
    16 x 1 = 16

    2 + 16 = 18 grams/mole

    2 x 4 = 8 18 x 4 = 72

    72/8 = 9

    9 x 0.112 = 1

    percentage yield = (0.745/1) x 100 = 74.5 %

    percentage yield = 74.5%
    You haven't gone wrong anywhere (or at least the mistake you made, your teacher also made). Simply, both of you have rounded part way through the question, meaning you got a lower answer than the precise one, and they got higher. Using unrounded values, I got 73.9%. Your method, however, is fine (and less confusing than your teacher's).
 
 
 
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