14.2 g C6H6 excess Br2 16.3 g C6H5Br

A chemist carries out the following reaction:

C6H6 + Br2 → C6H5Br + HBr

a. If he reacts 14.2 g of C6H6 with an excess of Br2,

what is the theoretical yield of C6H5Br?

b. If he isolated 16.3 g of C6H5Br,

what is his percent yield for this reaction?

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a. If he reacts 14.2 g of C6H6 with an excess of Br2,

what is the theoretical yield of C6H5Br?

excess Br2 이므로,

C6H6가 한계 반응물.

C6H6의 몰질량 = 78.11 g/mol 이므로,

14.2 g / (78.11 g/mol) = 0.182 mol C6H6

C6H6 + Br2 → C6H5Br + HBr

C6H6 : C6H5Br = 1 : 1 계수비(= 몰수비) 이므로,

생성되는 C6H5Br의 몰수 = 0.182 mol

C6H5Br의 몰질량 = 157.01 g/mol 이므로,

0.182 mol × (157.01 g/mol) = 28.6 g C6H5Br

---> theoretical yield of C6H5Br

b. If he isolated 16.3 g of C6H5Br,

what is his percent yield for this reaction?

percentage yield = (16.3 g / 28.6 g) × 100

= 57.0%

답: a. 28.6 g, b. 57.0%

[ 관련 예제 https://ywpop.tistory.com/19432 ]

한계 반응물과 수득률. 30.0 g C6H6 65.0 g Br2 42.3 g C6H5Br

[키워드] 한계 반응물 기준