Question

A chemist prepares hydrogen fluoride by means of the following reaction: CaF2 + H2SO4 —> CaSO4 + 2HF

The chemist uses 11g of CaF2 and an excess of H2SO4, and the reaction produces 2.22g of HF.
A) calculate the theoretical yield of HF
B) Calculate the percent yield of HF

pls helpppppp!!

Answer 1

a) Theoretical mass of HF from the reaction = 5.64 g

b) Percent yield of HF = 39.36%

Step-by-step explanation:

CaF2 + H2SO4 —> CaSO4 + 2HF

11g of CaF2 and an excess of H2SO4

This means that CaF2 is the limiting reagent. The limiting reagent determines how much products will be formed.

a) The theoretical yield of HF will be obtained from the stoichiometric balance of the reaction.

We first convert the 11g of CaF2 into number of moles.

Number of moles = (mass)/(molar mass)

Molar mass of CaF2 = 78.07 g/Lol

Number of moles of CaF2 = (11/78.07) = 0.141 moles.

From the balanced reaction,

1 mole of CaF2 gives 2 moles of HF

0.141 mole of CaF2 will give (2×0.141) mole of HF; that is, 0.282 moles of HF

Theoretical Mass of HF from the reaction

= (Number of moles of HF) × (Molar mass)

Molar mass of HF = 20.01 g/Lol

Theoretical mass of HF from the reaction

= 0.282 × 20.01 = 5.64 g

b) Percent yield = 100% × (Actual yield)/(Theoretical yield)

= 100% × (2.22/5.64)

= 39.36%

Hope this Helps!!!