What mass of sodium propylate can be obtained by reacting a mass of 25 g with sodium of a mass of 6.9.

Let’s find the amount of sodium substance by the formula:

n = m: M.

M (Na) = 23 g / mol.

n = 6.9 g: 23 g / mol = 0.3 mol.

Find the amount of propanol substance.

M (C3H7OH) = 12 × 3 + 8 + 16 = 60 g / mol.

n = 25 g: 60 g / mol = 0.42 mol.

Propanol is given in excess. We will solve the problem due to lack (sodium).

Let’s find the quantitative ratios of substances.

2СН3 – СН2- СН2ОН + 2Na → 2СН3 – СН2- СН2ОNа + H2 ↑.

For 2 mol of sodium, there are 2 mol of sodium propylate.

Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (Na) = n (C3H7OHa) = 0.3 mol.

Find the mass of sodium propylate.

m = n × M.

М (С3Н7ОNa) = 12 × 3 + 7 + 16 + 23 = 82 g / mol.

m = 82 g / mol × 0.3 mol = 24.6 g.

Answer: 24.6 g.