What mass of salt was formed as a result of neutralization of 400 g of sodium hydroxide with orthophosphate acid.

3NaOH + H3PO4 = Na3PO4 + 3H2O.
1) M (NaOH) = 23 + 16 + 1 = 40.
2) n (NaOH) = 400/40 = 10 mol.
3) n (Na3PO4) = 3 * 10 = 30 mol.
4) M (Na3PO4) = 23 * 3 + 31 + 16 * 4 = 164.
5) 30 * 164 = 4920g.
M is molar mass. It is calculated according to the periodic table by adding the atomic masses of all components of the substance.
n is the amount of substance. In this problem, it is found by dividing the mass by the molar mass.
The amount of the substance of the desired salt is determined by the reaction equation. For 3 mol of sodium hydroxide, there is one mol of sodium phosphate, which means the amount of the phosphate substance is three times the amount of the hydroxide substance.