Question

Из технического образца 2-бромбутана массой 30 г был получен алкен, который затем пропустили через слабощелочной раствор перманганата калия. Масса образовавшегося двухатомного спирта составила 9,54 г. Вычислите массовую долю примесей в исходном образце бромалкана, если известно, что выход реакции отщепления равен 90%, а выход реакции окисления-60%

Answer 1

C4H9Br + KOH=C4H8 + KBr +H2O
3C4H8 + 2KMnO4 + 4H2O=3C4H8(OH)2 + 2MnO2 + 2KOH
n(C4H8(OH)2)=9,54/90=0,106 моль
n(C4H8)=0,106/0,6=0,1767 моль
n(C4H9Br)=0,1767/0,9=0,1963 моль
m(C4H9Br)=0,1963*137=26,8931 г
w(примесей)=(30-26,8931)/30=10,356%~10,4%

Answer 2

Task: Calculation of Impurities in the Initial Sample of Bromobutane

To calculate the mass fraction of impurities in the initial sample of bromobutane, we need to consider the reaction steps and the given information.

1. The initial sample of 2-bromobutane with a mass of 30 g was used. 2. The 2-bromobutane was converted into an alkene. 3. The alkene was then treated with a weak alkaline solution of potassium permanganate. 4. The resulting product was a two-atom alcohol with a mass of 9.54 g. 5. The reaction yield for the cleavage reaction is 90%. 6. The reaction yield for the oxidation reaction is 60%.

To calculate the mass fraction of impurities, we need to determine the mass of the desired product (two-atom alcohol) and the mass of the impurities.

Let's calculate the mass of the desired product: - The mass of the two-atom alcohol is given as 9.54 g.

Now, let's calculate the mass of the impurities: - The reaction yield for the cleavage reaction is 90%, which means that 90% of the initial sample is converted into the desired product. Therefore, the mass of the impurities in the desired product is 10% of the mass of the desired product. - The reaction yield for the oxidation reaction is 60%, which means that 60% of the desired product is obtained from the alkene. Therefore, the mass of the impurities in the alkene is 40% of the mass of the desired product.

Using these calculations, we can determine the mass of the impurities in the initial sample of bromobutane.

Calculation:

Mass of the desired product (two-atom alcohol): 9.54 g

Mass of impurities in the desired product: - Cleavage reaction yield: 10% of 9.54 g = 0.954 g - Oxidation reaction yield: 40% of 9.54 g = 3.816 g

Total mass of impurities in the initial sample: - Impurities in the desired product: 0.954 g - Impurities in the alkene: 3.816 g

Therefore, the total mass of impurities in the initial sample of bromobutane is 0.954 g + 3.816 g = 4.77 g.

To calculate the mass fraction of impurities, we divide the mass of the impurities by the mass of the initial sample and multiply by 100:

Mass fraction of impurities = (4.77 g / 30 g) * 100 = 15.9%

Therefore, the mass fraction of impurities in the initial sample of bromobutane is approximately 15.9%.

Please note that the calculations provided are based on the given information and assumptions made about the reaction yields.