Question

При нитровании 0,5 моль фенола было получено 100 г. тринитрофенола.Определите пожалуйста практический выход (в %)от теоритически возможного=)

Answer 1

$C_6H_5OH+3HNO3=C_6H_2OH(NO_2)_3+3H_2O$

n(C6H5OH)=n(C6H2OH(NO2)3)=0.5моль

m(C6H2OH(NO2)3)=0.5*229=114.5г(практический)

W=100/114.5*100%=87.34%

Answer 2

Calculating Practical Yield of Trinitrophenol

To calculate the practical yield of trinitrophenol, we can use the formula:

Practical Yield (%) = (Actual Yield / Theoretical Yield) x 100%

Given: - Moles of phenol (n) = 0.5 mol - Mass of trinitrophenol obtained = 100 g - Molar mass of trinitrophenol = 229.11 g/mol

First, we need to calculate the theoretical yield of trinitrophenol.

The molar ratio between phenol and trinitrophenol is 1:1, so the moles of trinitrophenol produced will also be 0.5 mol.

The theoretical yield can be calculated using the formula:

Theoretical Yield = Moles of trinitrophenol x Molar mass of trinitrophenol

Theoretical Yield = 0.5 mol x 229.11 g/mol

Theoretical Yield = 114.55 g

Now, we can calculate the practical yield:

Practical Yield (%) = (100 g / 114.55 g) x 100%

Practical Yield ≈ 87.26%

Therefore, the practical yield of trinitrophenol is approximately 87.26%.