Respuesta :
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%.
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: The percentage yield of hydrogen fluoride is 80.3 %
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of [tex]CaF_2[/tex] = 5.95 kg = [tex]5.95\times 10^3g[/tex] (Conversion factor: 1 kg = 1000 g)
Molar mass of [tex]CaF_2[/tex] = 78.07 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }CaF_2=\frac{5.95\times 10^3g}{78.07g/mol}=76.21mol[/tex]
The chemical equation for the reaction of calcium fluoride and sulfuric acid follows:
[tex]CaF_2+H_2SO_4\rightarrow CaSO_4+2HF[/tex]
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 = [tex]\frac{2}{1}\times 76.21=152.42mol[/tex] 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:
[tex]152.42mol=\frac{\text{Mass of hydrogen fluoride}}{20g/mol}\\\\\text{Mass of hydrogen fluoride}=(152.42mol\times 20g/mol)=3048.4g=3.05kg[/tex]
To calculate the percentage yield of hydrogen fluoride, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of hydrogen fluoride = 2.45 kg
Theoretical yield of hydrogen fluoride = 3.05 kg
Putting values in above equation, we get:
[tex]\%\text{ yield of hydrogen fluoride}=\frac{2.45kg}{3.05kg}\times 100\\\\\% \text{ yield of hydrogen fluoride}=80.3\%[/tex]
Hence, the percentage yield of hydrogen fluoride is 80.3 %
