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

CaF2 + H2SO4 - > CaSO4 + 2HF

The chemist uses 11 g of CaFz 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) Theoretical Yield of HF = 5.64 grams

b) Percentage Yield = 39%

Explanation:

Reaction Given:

CaF2 + H2SO4 - > CaSO4 + 2HF

CaF2 = 11g

H2SO4 = Used in excess

HF = 2.2 g production = Actual Yield

So, Let's write down the molar masses:

Molar Mass of CaF2 = 78 g / mol

Molar Mass of HF = 20 g/mol

From the reaction, we can see the 1 mole of CaF2 gives the 2 moles of HF

i. e

a) Theoretical Yield of HF:

1 mole CaF2 = 2 moles HF

78 g CaF2 = 2 x 20 g of HF

78 g CaF2 = 40 g of HF

1 g CaF2 = 40g/78g of HF

And in the question it is given that chemist used 11 g of CaF2 so,

1 x 11 g of CaF2 = 11 x 40/78 g of HF

11 g of CaF2 = 440/78 g of HF

11 g of CaF2 = 5.64 g of HF

And this is the theoretical yield

Theoretical Yield of HF = 5.64 grams

b) Now, calculate the Percentage Yield of HF

Percentage Yield = Actual Yield / Theoretical Yield x 100

Percentage Yield = 2.2 g / 5.64 g x 100

Percentage Yield = 39%