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

mathrm{caf}_{2}+mathrm{h}_{2} mathrm{so}_{4}
ightarrow mathrm{caso}_{4}+2 mathrm{hf}

in one process, ( 5.95 mathrm{~kg} ) of ( mathrm{caf}_{2} ) is treated with an excess of ( mathrm{h}_{2} mathrm{so}_{4} ) and yields ( 2.35 mathrm{~kg} ) of ( mathrm{hf} ). calculate the percent yield of ( mathrm{hf} ). round your answer to 3 significant digits.

Explanation:

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

The molar mass of \(Ca\) is \(40.08\space g/mol\) and of \(F\) is \(19.00\space g/mol\). So, \(M_{CaF_2}=40.08+(2\times19.00)=78.08\space g/mol\)

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

Given \(m_{CaF_2} = 5.95\space kg=5950\space g\). Using \(n=\frac{m}{M}\), \(n_{CaF_2}=\frac{5950\space g}{78.08\space g/mol}\approx76.2\space mol\)

Step3: Determine moles of \(HF\) produced theoretically

From the balanced equation \(CaF_2 + H_2SO_4
ightarrow CaSO_4+2HF\), mole ratio \(CaF_2:HF = 1:2\). So, \(n_{HF}^{theoretical}=2\times n_{CaF_2}=2\times76.2 = 152.4\space mol\)

Step4: Calculate molar mass of \(HF\)

Molar mass of \(H = 1.01\space g/mol\) and \(F = 19.00\space g/mol\), so \(M_{HF}=1.01 + 19.00=20.01\space g/mol\)

Step5: Calculate theoretical mass of \(HF\)

Using \(m = n\times M\), \(m_{HF}^{theoretical}=152.4\space mol\times20.01\space g/mol\approx3050\space g = 3.05\space kg\)

Step6: Calculate percent yield

Percent yield formula is \(\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%\). Given actual yield \(m_{HF}^{actual}=2.35\space kg\). \(\text{Percent Yield}=\frac{2.35\space kg}{3.05\space kg}\times100\%\approx77.0\%\)

Answer:

\(77.0\%\)