# Saturated monohydric alcohol weighing 15 g reacted with metallic sodium. At the same time, hydrogen was released

**Saturated monohydric alcohol weighing 15 g reacted with metallic sodium. At the same time, hydrogen was released, the volume of which was 2.8 liters. Determine the formula for the alcohol and name it.**

1. Let’s make a diagram of the ongoing reaction:

2CnH2n + 1OH + 2Na → 2CnH2n + 1ONa + H2 ↑;

2.Calculate the chemical amount of released hydrogen:

n (H2) = V (H2): Vm = 2.8: 22.4 = 0.125 mol;

3.Let’s find the amount of the limiting monohydric alcohol:

n (CnH2n + 1OH) = n (H2) * 2 = 0.125 * 2 = 0.25 mol;

4.Calculate the relative molecular weight of alcohol:

Mr (CnH2n + 1OH) = m (CnH2n + 1OH): n (CnH2n + 1OH) = 15: 0.25 = 60;

5.determine the formula of alcohol by substituting the values of the relative atomic masses of the alcohol elements into the calculation of the molecular weight:

12n + 2n + 1 + 16 + 1 = 60;

14n + 18 = 60;

14n = 42;

n = 3.

Answer: propanol C3H7OH.