What mass of salt is formed by the interaction of 0.5 mol of sodium with 100 g of aminopropionic acid?

Let’s find the amount of aminopropionic acid substance by the formula:

n = m: M.

М (NH2 – CH2 –CH2 –COOH) = 89g / mol.

n = 100 g: 89 g / mol = 1.12 mol.

n (Na) = 0.5 mol.

The acid 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 salt.

Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (Na) = n (CH3 – CH2 – CH2ONa) = 0.5 mol.

Let’s find the mass of salt.

m = n × M.

M (CH3 CH2CH2ONa) = 82 g / mol.

m = 82 g / mol × 0.5 mol = 41 g.

Answer: 41 g.