Question

1)Сколько литров (н.у) этилена образуется из 4,6 г этилового спирта при 90%-ном выходе продукта?

Answer 1

C2H5OH=C2H4+H2O

Mr(C2H5OH)=46 грамм/моль

n(C2H5OH)=4,6/46=0,1 моль

Mr(C2H4)=28 грамм/моль

m(C2H4)=28*0,1=2,8 грамм - ответ, если бы выход равен 100%

w выхода (C2H4)=2,8*0,9=2,52 - ответ

Answer 2

Calculating Ethylene Production from Ethanol

To calculate the amount of ethylene produced from 4.6 grams of ethanol at a 90% yield, we can use the following steps:

1. Molar Mass of Ethanol (C2H5OH): - The molar mass of ethanol (C2H5OH) is 46.07 g/mol.

2. Molar Mass of Ethylene (C2H4): - The molar mass of ethylene (C2H4) is 28.05 g/mol.

3. Moles of Ethanol: - Calculate the moles of ethanol using the given mass and its molar mass: - Moles = Mass / Molar Mass - Moles = 4.6 g / 46.07 g/mol - Moles ≈ 0.1 mol

4. Moles of Ethylene Produced: - Ethanol undergoes dehydration to produce ethylene. The stoichiometry of this reaction is 1:1, meaning 1 mole of ethanol produces 1 mole of ethylene. - Since the yield is 90%, the moles of ethylene produced will be: - Moles of Ethylene = Moles of Ethanol * Yield - Moles of Ethylene = 0.1 mol * 0.9 - Moles of Ethylene ≈ 0.09 mol

5. Mass of Ethylene Produced: - Calculate the mass of ethylene produced using its molar mass: - Mass = Moles * Molar Mass - Mass = 0.09 mol * 28.05 g/mol - Mass ≈ 2.52 g

Answer:

Approximately 2.52 grams of ethylene (C2H4) are formed from 4.6 grams of ethanol (C2H5OH) at a 90% yield of the product.