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

CaF2 + H2SO4 → CaSO4 + 2HF

In one process, 5.75 kg of CaF₂ is treated with an excess of H₂SO₄ and yields 2.35 kg of HF. Calculate the percent yield of HF.

% yield

Answer 1

   CaF₂ + H₂SO₄ → CaSO₄ + 2HF

As excess of H₂SO₄ is used the limiting reactant would be CaF₂ is 5.75 g

Molar mass of CaF₂ = 78.07 g/mol

No. of moles of CaF₂ = Mass/molar mass = 5.75 g/78.07 g/mol = 0.074 kilomoles

Molar ratio of CaF₂ and HF = 1:2

So, moles of HF = 2 x 0.074 = 0.148 kilomoles

Molar mass of HF = 20.01 g/mol

Theoretical yield = Mass of HF = moles x molar mass = 0.148 kilomoles x 20.01 g/mol = 2.96 Kg

Actual yield = 2.35 Kg

Percent yield = (Actual yield/theoretical yield) x 100 = (2.35 Kg/2.96 Kg) x 100 = 79.39 %