Question
Asked Oct 30, 2019
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For the balanced equation Fe304 + 4H2 3Fe+ 4H20, if the reaction of 0.112 grams of H2 produces
0.745 grams of H20, what is the percent yield?
9.
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For the balanced equation Fe304 + 4H2 3Fe+ 4H20, if the reaction of 0.112 grams of H2 produces 0.745 grams of H20, what is the percent yield? 9.

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Expert Answer

Step 1

The balanced chemical equation is shown as,

Fe304+ 4 H2
3 Fe 4 H20
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Fe304+ 4 H2 3 Fe 4 H20

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Step 2

The percent yield is calculated based on the actual yield and theoretical yield. The actual yield of water is given as 0.745 grams. Hence to calculate theoretical yield of water, it is required to calculate the number of moles of water produced when 0.112 grams of hydrogen will react.

Given mass
No. of moles
Molecular Mass
0.112 g
No. of moles of hydrogen
0.056 mols
2
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Given mass No. of moles Molecular Mass 0.112 g No. of moles of hydrogen 0.056 mols 2

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Step 3

Hence the moles of water produced is 0.056 and ...

Mass No. of moles x Molecular Mass
Mass 0.056 x 18 = 1.008 grams
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Mass No. of moles x Molecular Mass Mass 0.056 x 18 = 1.008 grams

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