hydrogen fluoride is used in the manufacture of freons (which was banned in 2020, as it has been shown to…

hydrogen fluoride is used in the manufacture of freons (which was banned in 2020, as it has been shown to destroy ozone in our atmosphere), as well as in the production of aluminum metal. hf can be prepared by the following reaction: caf₂ + h₂so₄ → caso₄ + 2 hf. in one such process, 6.25 kg of calcium fluoride is treated with an excess of sulfuric acid and produces 2.85 kg of hf. what is the percent yield of hf during this process? report your answer to two significant figures.

hydrogen fluoride is used in the manufacture of freons (which was banned in 2020, as it has been shown to destroy ozone in our atmosphere), as well as in the production of aluminum metal. hf can be prepared by the following reaction: caf₂ + h₂so₄ → caso₄ + 2 hf. in one such process, 6.25 kg of calcium fluoride is treated with an excess of sulfuric acid and produces 2.85 kg of hf. what is the percent yield of hf during this process? report your answer to two significant figures.

Answer

Explanation:

Step1: Calculate the theoretical yield of HF

  • First, find the molar mass of (CaF_2). The molar mass of (Ca = 40.08\space g/mol) and (F=19.00\space g/mol). So, (M_{CaF_2}=40.08 + 2\times19.00=78.08\space g/mol).
  • The mass of (CaF_2) is (m_{CaF_2}=6.25\space kg = 6250\space g).
  • The number of moles of (CaF_2), (n_{CaF_2}=\frac{m_{CaF_2}}{M_{CaF_2}}=\frac{6250\space g}{78.08\space g/mol}\approx79.92\space mol).
  • From the balanced chemical equation (CaF_2 + H_2SO_4\rightarrow CaSO_4+2HF), the mole ratio of (CaF_2) to (HF) is (1:2).
  • The molar mass of (HF) is (M_{HF}=1.01 + 19.00=20.01\space g/mol).
  • The theoretical mass of (HF), (m_{HF}^{theo}=n_{CaF_2}\times2\times M_{HF}).
  • Substitute (n_{CaF_2}\approx79.92\space mol) and (M_{HF} = 20.01\space g/mol) into the formula: (m_{HF}^{theo}=79.92\space mol\times2\times20.01\space g/mol\approx3198\space g = 3.20\space kg).

Step2: Calculate the percent yield

  • The formula for percent yield is (\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100%).
  • The actual yield (m_{HF}^{act}=2.85\space kg) and the theoretical yield (m_{HF}^{theo} = 3.20\space kg).
  • (\text{Percent Yield}=\frac{2.85\space kg}{3.20\space kg}\times100% \approx 89%).

Answer:

(89%)