25.0 mL C6H6 18.0 mL C6H5NO2 percent yield

25.0 mL C6H6 18.0 mL C6H5NO2 percent yield

25.0 mL benzene 18.0 mL nitrobenzene percent yield

 

 

A student reacts 25.0 mL of benzene (C6H6, d 0.879, MW 78.0)

with a nitrating mixture contain excess nitric acid

to make 18.0 mL of nitrobenzene (C6H5NO2, d 1.204, MW 123.0)

according to the equation: C6H6 + HNO3 → C6H5NO2 + H2O

a) What is the percentage yield of the reaction?

b) What mass of C6H6 is left unreacted?

 

---------------------------------------------------

 

25.0 mL benzene의 몰수를 계산하면,

(25.0 mL) (0.879 g/mL) (1 mol / 78.0 g) = 0.2817 mol C6H6

 

 

 

excess nitric acid

---> HNO3 = 과잉 반응물, C6H6 = 한계 반응물.

( 참고: 한계 반응물 https://ywpop.tistory.com/3318 )

 

 

 

균형 맞춘 반응식

C6H6 + HNO3 → C6H5NO2 + H2O

 

C6H6 : C6H5NO2 = 1 : 1 계수비(= 몰수비) 이므로,

0.2817 mol C6H6 반응 시

생성되는 C6H5NO2의 몰수 = 0.2817 mol

 

 

 

0.2817 mol C6H5NO2의 부피를 계산하면,

(0.2817 mol) (123.0 g/mol) (1 mL / 1.204 g) = 28.8 mL C6H5NO2

---> 이론적 수득량

 

 

 

a) What is the percentage yield of the reaction?

수득율(%) = (18.0 mL / 28.8 mL) × 100 = 62.5%

( 참고 https://ywpop.tistory.com/61 )

 

 

 

b) What mass of C6H6 is left unreacted?

C6H6 = 한계 반응물이므로,

남아있는 C6H6의 질량 = 0 g.