Question

Пожалуйста, объясните как решать эту задачу и как понять последнее предложение(бромирование бензола произошло на 70%)? Буду очень благодарна.Газ, выделившийся при бромировании бензола массой 15,5 г, полностью прореагировал с водным раствором этиламина массой 30 г. Определите массовую долю этиламина в растворе, если бромирование бензола произошло на 70%

Answer 1

C6H6 + Br2 (FeBr3) = C6H5Br + HBr
C2H5NH2 + HBr = C2H5NH3Br
n(C6H6) = 15,5/78 = 0,199 моль
m(HBr) = 0,199*81 = 16,1 г
m(HBr)пр = 16,1*0,7 = 11,27 г
n(HBr) = 11,27/81 = 0,139 моль = n(C2H5NH2)
m(C2H5NH2) = 30 г
mв-ва(C2H5NH2) = 6,26 г
ω% = mв-ва*100%/mр-ра = 6,26*100/30 = 20,87%

это значит, что теоретически предполагалось одно количество бромбензола, а оказалось меньше. из-за каких-то потерь, например. в данном случае в результате первой реакции образовалось 70% от теоретически возможного количества.

Answer 2

Problem Explanation

The problem states that during the bromination of benzene, a gas was produced. This gas reacted completely with a solution of ethylamine, and the masses of the gas and ethylamine are given. The task is to determine the mass fraction of ethylamine in the solution, given that the bromination of benzene occurred to a 70% extent.

Solution

To solve this problem, we need to consider the stoichiometry of the reaction between the gas produced during bromination and ethylamine. From the given information, we know the mass of the gas produced and the mass of ethylamine that reacted.

Let's assume the gas produced is bromine gas (Br2), which is commonly used in bromination reactions. The balanced equation for the reaction between bromine and ethylamine is:

Br2 + 2C2H5NH2 → 2C2H5NHBr

According to the balanced equation, 1 mole of bromine gas reacts with 2 moles of ethylamine to produce 2 moles of ethylamine bromide.

To determine the mass fraction of ethylamine in the solution, we need to calculate the moles of ethylamine and the total moles of the solution.

Calculation

1. Calculate the moles of bromine gas produced: - Given mass of bromine gas = 15.5 g - Molar mass of bromine (Br2) = 159.808 g/mol - Moles of bromine gas = mass of bromine gas / molar mass of bromine gas

2. Calculate the moles of ethylamine: - Given mass of ethylamine = 30 g - Molar mass of ethylamine (C2H5NH2) = 45.083 g/mol - Moles of ethylamine = mass of ethylamine / molar mass of ethylamine

3. Calculate the total moles of the solution: - Moles of bromine gas reacted = moles of bromine gas produced (from step 1) - Moles of ethylamine bromide produced = 2 * moles of bromine gas reacted (from the balanced equation) - Total moles of the solution = moles of ethylamine + moles of ethylamine bromide produced

4. Calculate the mass fraction of ethylamine: - Mass fraction of ethylamine = (moles of ethylamine / total moles of the solution) * 100

Answer

By following the steps outlined above, you can calculate the mass fraction of ethylamine in the solution. Remember to use the given information that the bromination of benzene occurred to a 70% extent.

Let's calculate the values using the given data:

1. Moles of bromine gas produced: - Mass of bromine gas = 15.5 g - Molar mass of bromine (Br2) = 159.808 g/mol - Moles of bromine gas = 15.5 g / 159.808 g/mol

2. Moles of ethylamine: - Mass of ethylamine = 30 g - Molar mass of ethylamine (C2H5NH2) = 45.083 g/mol - Moles of ethylamine = 30 g / 45.083 g/mol

3. Total moles of the solution: - Moles of bromine gas reacted = Moles of bromine gas produced (from step 1) - Moles of ethylamine bromide produced = 2 * Moles of bromine gas reacted - Total moles of the solution = Moles of ethylamine + Moles of ethylamine bromide produced

4. Mass fraction of ethylamine: - Mass fraction of ethylamine = (Moles of ethylamine / Total moles of the solution) * 100

By substituting the values into the equations, you can calculate the mass fraction of ethylamine in the solution.