Question

Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction CaF2 + H2SO4 → CaSO4 + 2HF In one process, 6.25 kg of CaF2 is treated with an excess of H2SO4 and yields 2.35 kg of HF. Calculate the percent yield of HF. % yield

Answer 1

Answer: The percentage yield of HF is 73.36 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     ....(1)  

For calcium fluoride:

Given mass of calcium fluoride = 6.25 kg  = 6250 g   (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium fluoride = 78.07 g/mol

Putting values in above equation, we get:  

$\text{Moles of calcium fluoride}=\frac{6250g}{78.07g/mol}=80.05mol$

For the given chemical reaction:

$CaF_2+H_2SO_4\rightarrow CaSO_4+2HF$

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrofluoric acid

So, 80.05 moles of calcium fluoride will produce = $\frac{2}{1}\times 80.05=160.1mol$ of hydrofluoric acid

Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:

Moles of of hydrofluoric acid = 160.1 moles

Molar mass of hydrofluoric acid = 20.01 g/mol

Putting values in equation 1, we get:

$160.1mol=\frac{\text{Theoretical yield of hydrofluoric acid}}{20.01g/mol}=3203.6g=3.20kg$

To calculate the percentage yield of hydrofluoric acid, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of hydrofluoric acid = 2.35 kg

Theoretical yield of hydrofluoric acid = 3.20 kg

Putting values in above equation, we get:

$\%\text{ yield of hydrofluoric acid}=\frac{2.35g}{3.20g}\times 100\\\\\% \text{yield of hydrofluoric acid}=73.36\%$

Hence, the percentage yield of HF is 73.36 %