What mass of iron hydroxide (3) precipitates during the interaction of 400 g of sodium hydroxide with iron chloride (3).

Let’s compose the equation of the process:
3NaOH + FeCl3 = Fe (OH) 3 + 3NaCl – ion exchange, a precipitate of iron hydroxide is formed (3);

Let’s make the calculations:
M (NaOH) = 39.9 g / mol;

M Fe (OH) 3 = 106.8 g / mol;

Y (NaOH) = m / M = 400 / 39.9 = 10 mol.

Proportion:
10 mol (NaOH) – X mol Fe (OH) 3;

-3 mol -1 mol hence, X mol Fe (OH) 3 = 10 * 1/3 = 3.3 mol.

We find the mass of the sediment:
m Fe (OH) 3 = Y * M = 3.3 * 106.8 = 352.4 g

Answer: a precipitate of iron hydroxide weighing 352.4 g was obtained.