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.

A. The theoretical yield of HF is 5.64g

B. The percentage yield of HF is 39%

Explanation:

Step 1:

The balanced equation for the reaction:

CaF2 + H2SO4 - -> CaSO4 + 2HF

Step 2:

Determination of the mass of CaF2 that reacted and the mass of HF produced from the balanced equation. This is illustrated below:

Molar Mass of CaF2 = 40 + (19x2) = 40 + 38 = 78g/mol

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF from the balanced equation = 2 x 20 = 40g.

From the balanced equation above,

78g of CaF2 reacted and 40g of HF were produced.

A. Determination of the theoretical yield of HF.

This is illustrated below:

From the balanced equation above,

78g of CaF2 reacted to produce 40g of HF.

Therefore, 11g of CaF2 will react to produce = (11 x 40) / 78 = 5.64g of HF.

The theoretical yield of HF is 5.64g

B. Determination of the percentage yield.

The percentage yield of HF can be obtained as follow:

Actual yield = 2.2g

Theoretical yield = 5.64g

Percentage yield = ?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 2.2/5.64 x 100

Percentage yield = 39%

The percentage yield of HF is 39%

39.3%

Explanation:

Our guide in solving the problem must be the reaction equation hence it is pertinent to put down first:

CaF2 + H2SO4 - -> CaSO4 + 2HF

We have a very important information in the question, sulphuric acid is present in excess. This implies that calcium fluoride is the limiting reactant.

Number of moles of calcium fluoride reacted = mass of calcium fluoride reacted / molar mass of calcium fluoride

Molar mass of calcium fluoride = 78.07 g/mol

Number of moles of calcium fluoride = 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 0.28 moles of hydrogen fluoride = number of moles * molar mass

Molar mass of hydrogen flouride = 20.01 g/mol

Mass of HF = 0.28 moles * 20.01 g/mol = 5.6 g this is the 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%