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+> 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(> ch3ch2och3(l)+h2o(l)
calculate the percent yield if 60.0g of ethanol reacts to produce 13.0g of ether % yield?

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

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%

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g