Question

In the reaction of 53.80 g of nitric acid with 74.88 g of benzene, calculate the percent yield if 65.62 g of nitrobenzene is obtained in the lab. Report your answer to the tenths of a percent without units.

Answer 1

Molar mass of C6H6,

MM = 6*MM(C) + 6*MM(H)

= 6*12.01 + 6*1.008

= 78.108 g/mol

mass(C6H6)= 74.88 g

use:

number of mol of C6H6,

n = mass of C6H6/molar mass of C6H6

=(74.88 g)/(78.11 g/mol)

= 0.9587 mol

Molar mass of HNO3,

MM = 1*MM(H) + 1*MM(N) + 3*MM(O)

= 1*1.008 + 1*14.01 + 3*16.0

= 63.018 g/mol

mass(HNO3)= 53.8 g

use:

number of mol of HNO3,

n = mass of HNO3/molar mass of HNO3

=(53.8 g)/(63.02 g/mol)

= 0.8537 mol

Balanced chemical equation is:

C6H6 + HNO3 ---> C6H5NO2 + H2O

1 mol of C6H6 reacts with 1 mol of HNO3

for 0.9587 mol of C6H6, 0.9587 mol of HNO3 is required

But we have 0.8537 mol of HNO3

so, HNO3 is limiting reagent

we will use HNO3 in further calculation

Molar mass of C6H5NO2,

MM = 6*MM(C) + 5*MM(H) + 1*MM(N) + 2*MM(O)

= 6*12.01 + 5*1.008 + 1*14.01 + 2*16.0

= 123.11 g/mol

According to balanced equation

mol of C6H5NO2 formed = (1/1)* moles of HNO3

= (1/1)*0.8537

= 0.8537 mol

use:

mass of C6H5NO2 = number of mol * molar mass

= 0.8537*1.231*10^2

= 1.051*10^2 g

% yield = actual mass*100/theoretical mass

= 65.62*100/1.051*10^2

= 62.43%

Answer: 62.4