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In order to etch a piece of glass, a glassmaker needs to use hydrogen fluoride....

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In order to etch a piece of glass, a glassmaker needs to use hydrogen fluoride. Hydrogen and fluorine react to produce this compound. What is the percent yield for this reaction if 57.0 grams of fluorine react with an excess of hydrogen to actually produce 45.0 Grams of HF?

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jerichorayel's profile pic

Posted (Answer #2)

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The reaction of hydrogen and fluorine gases can be written as:

`H_2 + F_2 -> 2 HF`

In getting the percent yield of the reaction, we can use the expression:

`percent yield = (actual yield)/(theo retical yield) * 100`

actual yield = 45.0

theoretical yield  = ?

 

We can obtain the theoretical yield of HF by stoichiometry.

`57.0 grams F_2 * (1 mol e F_2)/(37.997 grams F_2) * (2 mol es HF)/(1 mol e F_2) * (20.01 grams HF)/(1 mol e HF)`

= 60.03 grams HF are produced theoretically. 

 

`percent yield = (actual yield)/(theoretical yield) * 100`

`percent yield = (45.0)/(60.03) * 100`

percent yield = 75.0 % 

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mvcdc's profile pic

Posted (Answer #1)

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H2 + F2 --> 2HF

MW (F2) = 19*2 = 38 grams/mole

MW (HF) = 1 + 19 = 20 grams/mole

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57.0 grams F2 x (1 mole F2/ 38 grams F2) x (2 moles HF / mole F2) x (20 grams HF / 1 mole HF) = 60 grams HF.

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Theoretical Yield = 60 grams

Actual Yield = 45 grams

%Yield = Actual/Theoretical = 45/60 x 100 = 75%

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