65.3 g C2H5OH give 58.2 g C4H10O percent yield

65.3 g C2H5OH give 58.2 g C4H10O percent yield

65.3 g ethyl alcohol give 58.2 g diethyl ether percent yield

 

 

Ethyl alcohol can be converted to diethyl ether (an anaesthetic) by the reaction:

2C2H5OH → C4H10O + H2O

Some of the diethyl ether product is lost due to evaporation.

If 65.3 g of ethyl alcohol yields 58.2 g of diethyl ether,

calculate the percentage yield.

 

If 65.3 g of ethyl alcohol gives 58.2 g of diethyl ether,

what is the percentage yield?

 

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C2H5OH의 몰질량 = 46.07 g/mol 이므로,

65.3 g / (46.07 g/mol) = 1.4174 mol C2H5OH

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

2C2H5OH → C4H10O + H2O

C2H5OH : C4H10O = 2 : 1 계수비(= 몰수비) 이므로,

1.4174 mol C2H5OH 반응 시, 생성되는 C4H10O의 몰수를 계산하면,

C2H5OH : C4H10O = 2 : 1 = 1.4174 mol : ? mol

 

? = 1.4174 / 2 = 0.7087 mol C4H10O

 

 

 

C4H10O의 몰질량 = 74.12 g/mol 이므로,

0.7087 mol × (74.12 g/mol) = 52.5288 g C4H10O

---> 이론적 수득량

 

 

 

percentage yield를 계산하면,

( 참고: 수득률 https://ywpop.tistory.com/61 )

 

(58.2 g / 52.5288 g) × 100 = 110.8%

---> 실험 엉터리로 했네.

( 참고: 제대로한 실험 https://ywpop.tistory.com/20775 )