What is Percent Yeild of reaction

Question

0.428 liters of hydrogen gas was reacted with excess fluorine gas to produce hydrogen fluoride. Given that the density of the hydrogen gas is 0.082 g/L, what was the percent yield of the reaction if 0.646645537425743 grams of hydrogen fluoride was actually produced?

Answer 1

Hello Matt

 

First let's write the balanced chemical reaction:

 

H₂ (g) + F₂ (g) → 2HF (g)

 

Next, we need to determine the theoretical yield of HF in grams using the following stoichiometry process:

 

L H₂ → g H₂ → mol H₂ → mol HF → g HF

 

Molar masses needed for the stoichiometry process are: H₂ = 2 g/mol,  HF = 20 g/mol.  From the chemical reaction, you can see that 1 mole of H₂ forms 2 moles of HF.

1) Convert volume of H₂ gas from Liters to grams using the given density.

 

m = d x V = 0.082 g/L x 0.428 L = 0.0351 g H₂

 

2) Convert grams of H₂ to grams of HF to obtain the theoretical yield.

 

0.0351 g H₂ x (1 mol H₂ / 2 g H₂) x (2 mol HF / 1 mol H₂) x (20 g HF / 1 mol HF) = 0.702 g HF

 

3) %-Yield = (Actual yield / Theoretical yield ) x 100%

 

Actual yield = 0.646645537425743 g

 

%-Yield = (0.646645537425743 g / 0.702 g) x 100% = 92.1% (to the correct # of significant figures)