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, 6.25kg of CaF2 is treated with an excess of H2SO4 and yields 2.35 kg of HF. Calculate the percent yield of HF.
(% yield)
2) Diethyl ether is produced from ethanol according to the following equation. 2CH3CH2OH(l)---->CH3CH2OCH3(l)+H2O(l)
Calculate the percent yield if 60.0g of ethanol reacts to produce 13.0g of ether % yield?

Answer 1

Yield(HF)=73.35% Yield(CH3CH2OCH2CH3)=27.69%

Step-by-step explanation:

CaF2+H2SO4---->CaSO4+2HF

n(CaF2)=m(CaF2)/M(CaF2)=6250/78.08=80.05 mol

n(HF)=m(HF)/M(HF)=2350/20.01=117.44 mol

We obtain two moles of HF from one mole CaF2. So the yield will be

Yield(HF)=n(HF)/(2*n(CaF2))=73.35%

2CH3CH2OH(l)---->CH3CH2OCH3(l)+H2O(l)

n(CH3CH2OH)=m(CH3CH2OH)/M(CH3CH2OH)= 60/46.07=1.3 mol

n(CH3CH2OCH2CH3)=m(CH3CH2OCH2CH3)/M(CH3CH2OCH2CH3)=13/74.12=0.18 mol

We obtain one mole of diethyl ether from two molecules of ethanol so the yield will be

Yield(CH3CH2OCH2CH3)=n(CH3CH2OCH2CH3)*2/n(CH3CH2OH)=27.69%