14.2 g C6H6 excess Br2 16.3 g C6H5Br yield

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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> C6H6의 몰질량 = 78.11 g/mol

> C6H5Br의 몰질량 = 157.01 g/mol

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

what is the theoretical yield of C6H5Br?

C6H6 : C6H5Br

= 78.11 g : 157.01 g = 14.2 g : ? g

? = 157.01 × 14.2 / 78.11

= 28.5 g C6H5Br

( 참고: 질량비로 계산 https://ywpop.tistory.com/22115 )

b. If he isolated 16.3 g of C6H5Br,

what is his percent yield for this reaction?

수득률 = (실제 수득량 / 이론적 수득량) × 100

= (16.3 g / 28.5 g) × 100

= 57.2%

( 참고 https://ywpop.tistory.com/61 )

[ 관련 예제 https://ywpop.tistory.com/19432 ] 벤젠 30.0 g과 브로민 65.0 g 반응

[키워드] C6H6 + Br2 → C6H5Br + HBr 기준