Question

14 граммов углеводорода этиленового ряда, полученного из предельного одноатомного спирта, реагирует с 80 гр брома. Вычислите молярную массу взятого спирта. М,(Bry)=160 Дам много баллов пж помните

Answer 1

Ответ:

С2H4 - этилен

Объяснение:

Из предельных спиртов через дегидратацию получают углеводороды алкенового ряда, которые присоединяют бром в полном объеме по месту разрыва двойной связи. Поэтому реакция идет в мольном отношении 1 к 1.

n(Br2)=80/160=0.5 моль

Значит и алкена было 0,5 моль

M(CxHy)=14/0.5=28 и это этилен.

12*n+2n=28

14n=28

n=2

C2H4

Answer 2

To calculate the molar mass of the alcohol, we need to determine the number of moles of bromine that reacted with the given amount of the ethylene series hydrocarbon.

Let's start by finding the number of moles of bromine that reacted. We can use the formula:

moles = mass / molar mass

Given that the mass of bromine is 80 g and the molar mass of bromine is 160 g/mol, we can calculate the number of moles of bromine:

moles of bromine = 80 g / 160 g/mol = 0.5 mol [[1]]

Since the reaction is a one-to-one ratio between the ethylene series hydrocarbon and bromine, we can conclude that 0.5 moles of the ethylene series hydrocarbon reacted.

Now, let's calculate the molar mass of the ethylene series hydrocarbon. We know that 14 grams of the hydrocarbon reacted with 0.5 moles of bromine.

molar mass = mass / moles

Given that the mass of the hydrocarbon is 14 g and the number of moles is 0.5 mol, we can calculate the molar mass:

molar mass of the hydrocarbon = 14 g / 0.5 mol = 28 g/mol

Therefore, the molar mass of the taken alcohol is 28 g/mol.

I hope this helps! Let me know if you have any further questions.

[[1]] Source: [Chemistry LibreTexts: Molar Mass](https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/02._Atoms%2C_Molecules%2C_and_Ions/2.6%3A_Formula_Mass_and_the_Mole_Concept)