Question

При действии избытка брома на бензол химическим количеством 0,2 моль получили бромбензол массой 29,83 г. Найдите массовую долю выхода бромбензола от теоретического ( в % )

Answer 1

0,2 моль                       х г

C6H6    +    Br2    =   C6H5Br  +  HBr

 1 моль                 157 г/моль

 

х = 157*0,2 = 31,4 г.   Это теоретический выход бромбензола.

Реальный 29,83 г.

Массовая доля выхода в % от теоретического:

29,83*100%/31,4 = 95 %

Ответ: 95 %.

Answer 2

Calculation of the Mass Fraction of Bromobenzene

To find the mass fraction of bromobenzene, we need to calculate the theoretical yield of bromobenzene and compare it to the actual yield.

Given: - Chemical quantity of bromine used: 0.2 moles - Mass of bromobenzene obtained: 29.83 grams

To calculate the theoretical yield of bromobenzene, we need to determine the molar mass of bromobenzene. The molar mass of bromobenzene is the sum of the atomic masses of its constituent elements.

The molar mass of bromine (Br) is approximately 79.9 g/mol, and the molar mass of benzene (C6H6) is approximately 78.11 g/mol.

Therefore, the molar mass of bromobenzene (C6H5Br) is: 6 * (12.01 g/mol) + 5 * (1.01 g/mol) + 79.9 g/mol = 157.01 g/mol

Now, we can calculate the theoretical yield of bromobenzene using the stoichiometry of the reaction. The balanced chemical equation for the reaction between bromine and benzene to form bromobenzene is:

C6H6 + Br2 -> C6H5Br + HBr

From the balanced equation, we can see that the stoichiometric ratio between bromine and bromobenzene is 1:1. This means that for every 1 mole of bromine used, 1 mole of bromobenzene is produced.

Since the chemical quantity of bromine used is 0.2 moles, the theoretical yield of bromobenzene is also 0.2 moles.

To convert the theoretical yield from moles to grams, we can use the molar mass of bromobenzene:

Theoretical yield of bromobenzene = 0.2 moles * 157.01 g/mol = 31.402 grams

Now, we can calculate the mass fraction of bromobenzene by dividing the actual yield by the theoretical yield and multiplying by 100:

Mass fraction of bromobenzene = (Actual yield / Theoretical yield) * 100

Mass fraction of bromobenzene = (29.83 g / 31.402 g) * 100 ≈ 95.0%

Therefore, the mass fraction of bromobenzene from the theoretical yield is approximately 95.0%.

Please note that the calculated mass fraction is an approximation and may vary depending on the experimental conditions and purity of the reactants used.