Question

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.

Explanation:

Step1: Calculate the molar mass of \(CaF_2\) and \(HF\)

The molar mass of \(CaF_2\): \(M_{CaF_2}=40.08+(2\times19.00)=78.08\space g/mol\)
The molar mass of \(HF\): \(M_{HF}=1.01 + 19.00=20.01\space g/mol\)

Step2: Convert the mass of \(CaF_2\) to moles

Given \(m_{CaF_2}=6.25\space kg = 6250\space g\)
\(n_{CaF_2}=\frac{m_{CaF_2}}{M_{CaF_2}}=\frac{6250\space g}{78.08\space g/mol}\approx79.92\space mol\)

Step3: Calculate the theoretical yield of \(HF\)

From the balanced equation \(CaF_2 + H_2SO_4
ightarrow CaSO_4+2HF\), the mole ratio of \(CaF_2\) to \(HF\) is \(1:2\)
So, \(n_{HF}(theoretical) = 2\times n_{CaF_2}=2\times79.92\space mol = 159.84\space mol\)
\(m_{HF}(theoretical)=n_{HF}(theoretical)\times M_{HF}=159.84\space mol\times20.01\space g/mol\approx3198.4\space g = 3.1984\space kg\)

Step4: Calculate the percent yield

Given \(m_{HF}(actual)=2.85\space kg\)
Percent yield \(=\frac{m_{HF}(actual)}{m_{HF}(theoretical)}\times100\%=\frac{2.85\space kg}{3.1984\space kg}\times100\%\approx89\%\)

Answer:

\(89\%\)