28.4 g of 5-valence phosphorus oxide was treated with sodium hydroxide 100 g. Determine the mass of the salt.

To solve the problem, write down the equation:
1. P2O5 + 6NaOH = 2Na3PO4 + 3H2O – ion exchange reaction, sodium phosphate was obtained;
2. We make calculations:
M (P2O5) = 141.8 g / mol;
M (NaOH) = 39.9 g / mol;
M (Na3PO4) = 163.6 g / mol.
2. Determine the number of moles of the starting materials, if the mass is known:
Y (P2O5) = m / M = 28.4 / 141.8 = 0.2 mol (deficient substance);
Y (NaOH) = m / M = 100 / 39.9 = 2.5 mol (in excess).
3. Proportion:
0.2 mol (P2O5) – X mol (Na3PO4);
-1 mol – 2 mol hence, X mol (Na3PO4) = 0.2 * 2/1 = 0.4 mol.
4. Find the mass of salt:
m (Na3PO4) = 0.4 * 163.6 = 65.44 g.
Answer: in the course of the reaction, sodium phosphate with a mass of 65.44 g was obtained.