In the reaction of an unknown saturated monohydric alcohol weighing 17.92 g with aluminum, hydrogen

In the reaction of an unknown saturated monohydric alcohol weighing 17.92 g with aluminum, hydrogen was released with a volume of 6.272 dm3. Determine the formula for the alcohol.

According to the condition of the problem, we write down the general formula for the limiting monohydric alcohol: CnH2n + 1OH.
Let’s compose the reaction equation schematically:
6ROH + 2Al = 3H2 + 2Al (RO) 3 – substitution reaction, hydrogen gas is released;
Let’s convert units of volume of hydrogen to liters:
1 l = 1 dm3; 6.272 dm3 = 6.272 l;
Let’s calculate the amount of moles of hydrogen, alcohol:
1 mol of gas at n. y – 22.4 l;
X mol (H2) – 6.272 liters. hence, X mol (H2) = 1 * 6.272 / 22.4 = 0.28 l;
X mol (ROH) – 0.28 mol (H2);
-6 mol – 3 mol hence, X mol (ROH) = 6 * 0.28 / 3 = 0.56 mol;
We find the molar mass of alcohol by the formula:
M (ROH) = m / Y = 17.92 / 0.56 = 32 g / mol;
We derive the molecular formula of alcohol:
CnH2n + 1OH = 32 g / mol;
12n + 2 + 1 + 16 + 1 = 32;
12n = 20;
n = 1.6;
Molecular formula – CH3OH (methanol), M (CH3OH) = 12 + 3 + 16 + 1 = 32 g / mol. The problem was solved correctly.
Answer: CH3OH – alcohol methanol.