Question

A 15.6g sample of C6H6 reacts with an excess of HNO3. After the laboratory process, 18.0g of C6H5NO2 are obtained, what is the% yield of the reaction?

C6H6 + HNO3 → C6H5NO2 + H2O

Answer 1

Molar mass of C6H6 = ( 6 12.01 ) + ( 6 1.0079 ) = 78.11 g / mol

Molar mass of C6H5NO2 = ( 6 12.01 ) + ( 5 1.0079 ) + 14.0067 + ( 2 16.00 ) = 123.11 g / mol

We have relation, no. of moles = mass / molar mass

No. of moles of C6H6 consumed in the reaction = 15.6 g / ( 78.11 g /mol ) = 0.1997 mol

Consider reaction, C6H6 + HNO3 C6H5NO2 + H2O

According to reaction, 1 mol C6H5NO2 is produced from 1 mol C6H6. Hence, moles of  C6H5NO2 produced from 0.1997 mol C6H6 would be

0.1997 mol C6H6 ( 1 mol C6H5NO2 / 1 mol C6H6 ) = 0.1997 mol C6H5NO2

Mass of C6H5NO2 ​​​​​​​produced in the reaction = 0.1997 mol 123.11 g / mol = 24.58 g

Theoretical yield of  C6H5NO2 ​​​​​​​= 24.58 g

% yield of reaction = ( Actual yield of product / Theoretical yield of product ) 100

% yield of reaction = (18.0 g / 24.58 g )   100 = 73.22 %

ANSWER : % yield of reaction = 73.22 %