When propanol interacts with metallic sodium, 0.01 mol of hydrogen is released.

When propanol interacts with metallic sodium, 0.01 mol of hydrogen is released. Determine the mass of the alcohol that has reacted.

To solve the problem, it is necessary to compose the reaction equation:
2С3Н7ОН + 2Na = 2C3H7ONa + H2 – substitution reaction, hydrogen is released;
M (C3H7OH) = 3 * 12 + 8 + 16 = 60 g / mol;
M (H2) = 2 g / mol;
Let’s make a proportion according to the equation:
0.01 mol (H2) – X mol (C3H7OH);
1 mol -2 mol from here, X mol (C3H7OH) = 0.01 * 2/1 = 0.02 mol;
We find the mass of propanol by the formula:
m (C3H7OH) = Y * M = 0.02 * 60 = 1.2 g.
Answer: propanol with a mass of 1.2 g enters the reaction.