Calculate the mass of ethyl alcohol if 28 liters of ethylene were obtained during dehydration.

Let’s implement the solution:
1. According to the condition of the problem, we write down the equation:

С2Н5ОН = Н2О + С2Н4 – dehydration, ethylene was obtained;

2. Let’s make the calculations:
M (CH5OH) = 46 g / mol;
M (C2H4) = 28 g / mol.
3. Determine the number of moles of the product:
1 mol of gas at normal level – 22.4 liters;
X mol (C2H4) – 28 L. hence, X mol (C2H4) = 1 * 28 / 22.4 = 1.25 mol;
Y (C2H5OH) = 1.25 mol, since the amount of substances according to the equation is 1 mol.
4. Find the mass of alcohol:
m (C2H5OH) = Y * M = 1.25 * 46 = 57.5 g.
Answer: to carry out the process, you will need ethyl alcohol weighing 57.5 g.