Question

In one process, 5.95 kg of caf2 is treated with an excess of h2so4 and yields 2.45 kg of hf. calculate the percent yield of hf.

Answer 1

Answer is: yield of reaction is 80,3%.
Chemical reaction: CaF₂ + H₂SO₄ → CaSO₄ + 2HF.
m(CaF₂) = 5,95 kg · 1000 g/kg = 5950 g.
n(CaF₂) = m(CaF₂) ÷ M(CaF₂).
n(CaF₂) = 5950 g ÷ 78 g/mol.
n(CaF₂) = 76,28 mol.
From chemical reaction: n(CaF₂) : n(HF) = 1 : 2.
n(HF) = 76,28 mol · 2 = 152,56 mol.
m(HF) = 152,56 mol · 20 g/mol.
m(HF) = 3051,2 g ÷ 1000 g/kg = 3,0512 kg.
yield = 2,45 kg kg ÷ 3,0512 kg · 100% = 80,3%.

Answer 2

Answer: The percentage yield of hydrogen fluoride is 80.3 %

Explanation:

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

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

Given mass of $CaF_2$ = 5.95 kg = $5.95\times 10^3g$    (Conversion factor:  1 kg = 1000 g)

Molar mass of $CaF_2$ = 78.07 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CaF_2=\frac{5.95\times 10^3g}{78.07g/mol}=76.21mol$

The chemical equation for the reaction of calcium fluoride and sulfuric acid follows:

$CaF_2+H_2SO_4\rightarrow CaSO_4+2HF$

As, sulfuric acid  is present in excess. It is considered as an excess reagent.

Calcium fluoride is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrogen fluoride

So, 76.21 moles of calcium fluoride will produce = $\frac{2}{1}\times 76.21=152.42mol$ of hydrogen fluoride

Now, calculating the mass of hydrogen fluoride by using equation 1, we get:

Molar mass of hydrogen fluoride = 20 g/mol

Moles of hydrogen fluoride = 152.42 moles

Putting values in equation 1, we get:

$152.42mol=\frac{\text{Mass of hydrogen fluoride}}{20g/mol}\\\\\text{Mass of hydrogen fluoride}=(152.42mol\times 20g/mol)=3048.4g=3.05kg$

To calculate the percentage yield of hydrogen fluoride, we use the equation:

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

Experimental yield of hydrogen fluoride = 2.45 kg

Theoretical yield of hydrogen fluoride = 3.05 kg

Putting values in above equation, we get:

$\%\text{ yield of hydrogen fluoride}=\frac{2.45kg}{3.05kg}\times 100\\\\\% \text{ yield of hydrogen fluoride}=80.3\%$

Hence, the percentage yield of hydrogen fluoride is 80.3 %