A chemist prepares hydrogen fluoride by means of the following reaction:

CaF2 + H2SO4 --> CaSO4 + 2HF

The chemist uses 11 g of CaF2 and an excess of H2SO4, and the reaction produces 2.2 g of HF.
(a) Calculate the theoretical yield of HF.

(b) Calculate the percent yield of HF.

39.3%

Explanation:

CaF2 + H2SO4 --> CaSO4 + 2HF

We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:

For CaF2;

Number of moles reacted= mass/molar mass

Molar mass of CaF2= 78.07 g/mol

Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride

Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride

0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride

Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass

Molar mass of hydrogen fluoride= 20.01 g/mol

Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)

Actual yield of HF was given in the question as 2.2g

% yield of HF= actual yield/ theoretical yield ×100

%yield of HF= 2.2/5.6 ×100

% yield of HF= 39.3%

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