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# percentage yield watch

1. 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%
2. (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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Updated: April 14, 2013
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