Question

Помогите !!!!!!! При действии избытка брома на 0,2 моль бензола получили 29.83 г бромбензолу. Определите массовое долю выхода бромбензолу от теоретического (в%).

Answer 1

C₆H₆ + Br₂ → C₆H₅-Br + HBr --( катализатор FeBr₃)
n(C₆H₆) = n(C₆H₅-Br)
m(C₆H₅-Br) = 0,2 моль * 157 г/моль = 31,4 г. -- теоретический выход бромбензола
η = m(практ)/m(теорет) * 100% = 29,83 г/31,4 г. *100% = 95%

Answer 2

Calculation of the Mass Fraction of Bromobenzene

To determine the mass fraction of bromobenzene, we need to compare the actual yield of bromobenzene to the theoretical yield. The theoretical yield is the maximum amount of product that can be obtained based on the stoichiometry of the reaction.

Given: - Excess bromine: 0.2 moles - Mass of bromobenzene obtained: 29.83 g

To calculate the mass fraction, we need to find the theoretical yield of bromobenzene first. 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 benzene and bromobenzene is 1:1. This means that for every mole of benzene reacted, we should obtain one mole of bromobenzene.

Since we have an excess of bromine, the limiting reactant is benzene. Therefore, the number of moles of bromobenzene obtained is equal to the number of moles of benzene reacted.

Using the molar mass of bromobenzene (157.02 g/mol), we can calculate the theoretical yield of bromobenzene:

Theoretical yield = moles of bromobenzene * molar mass of bromobenzene

To find the moles of bromobenzene, we can use the molar mass of bromobenzene and the mass of bromobenzene obtained:

moles of bromobenzene = mass of bromobenzene obtained / molar mass of bromobenzene

Let's calculate the moles of bromobenzene:

moles of bromobenzene = 29.83 g / 157.02 g/mol = 0.1899 mol

Now, we can calculate the mass fraction of bromobenzene:

Mass fraction of bromobenzene = (actual yield / theoretical yield) * 100

Substituting the values:

Mass fraction of bromobenzene = (0.1899 mol / 0.2 mol) * 100 = 94.95%

Therefore, the mass fraction of bromobenzene, relative to the theoretical yield, is approximately 94.95%.

Please note that this calculation assumes that the reaction proceeded to completion and that there were no side reactions or losses during the process.