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Chapter 3: Problem 89

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 $$ \mathrm{CaF}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{CaSO}_{4}+2 \mathrm{HF} $$ In one process \(6.00 \mathrm{~kg}\) of \(\mathrm{CaF}_{2}\) are treated with an excess of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and yield \(2.86 \mathrm{~kg}\) of \(\mathrm{HF}\). Calculate the percent yield of HF.

Short Answer

Expert verified
The percent yield of Hydrogen Fluoride (HF) is approximately 93.1 %

Step by step solution

01

Calculate Moles of \(CaF_2\)

First, convert the mass of \(CaF_2\) used to moles. According to the Periodic Table, the molar mass of \(CaF_2\) (Calcium is 40.08 g/mol, Fluorine is 18.998 g/mol x 2) is 40.08 g/mol + (18.998 g/mol x 2) = 78.076 g/mol. Using the formula for moles (moles = mass / molar mass), the moles of \(CaF_2\) used = mass \(CaF_2\ / molar mass = 6000 g / 78.076 g/mol = 76.8 moles\)
02

Calculate Theoretical Yield of HF

Using the balanced chemical equation, it can be seen that each mole of \(CaF_2\) produces 2 moles of HF. Therefore, the theoretical yield of HF in moles = 2 * moles of \(CaF_2\) used = 2 * 76.8 moles = 153.6 moles. Then, convert moles of HF to mass (in g) using the formula mass = moles * molar mass. The molar mass of HF (Hydrogen is 1.007 g/mol, Fluorine is 18.998 g/mol) is 1.007 g/mol + 18.998 g/mol = 20.005 g/mol. Therefore, the theoretical yield of HF = moles HF * molar mass = 153.6 moles * 20.005 g/mol = 3072 g or 3.072 kg.
03

Calculate Percent Yield

The percent yield is calculated by dividing the actual yield by the theoretical yield and then multiplying by 100. Percent yield = (actual yield / theoretical yield) * 100 = (2.86 kg / 3.072 kg) * 100 = 93.1 %.

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