Phosphorus (V) oxide, obtained by oxidation of 31 g of phosphorus, is dissolved in 495 g of water

Phosphorus (V) oxide, obtained by oxidation of 31 g of phosphorus, is dissolved in 495 g of water to form orthophosphoric acid. 44.8 liters of ammonia were passed into this solution. Determine the composition of the obtained salt and its concentration in the solution (mass fraction in percent).

Given: m (P) = 31 g m (H2O) = 495 g V (NH3) = 44.8 l
Find: ω (salt) -?
Decision:
1) Write the reaction equations: 4P + 5O2 => 2P2O5; P2O5 + 3H2O => 2H3PO4;
2) Find the amount of substance P: n (P) = m (P) / Mr (P) = 31/31 = 1 mol;
3) Find the amount of substance P2O5 (taking into account the reaction equation): n (P2O5) = n (P) / 2 = 1/2 = 0.5 mol;
4) Find the mass of the substance P2O5: m (P2O5) = n (P2O5) * Mr (P2O5) = 0.5 * 142 = 71 g;
5) Find the amount of the substance H3PO4 (taking into account the reaction equation): n (H3PO4) = n (P2O5) * 2 = 0.5 * 2 = 1 mol;
6) Find the amount of substance NH3: n (NH3) = V (NH3) / Vm = 44.8 / 22.4 = 2 mol;
7) Find the mass of the substance NH3: m (NH3) = n (NH3) * Mr (NH3) = 2 * 17 = 34 g;
8) Write the equation of the salt formation reaction: the ratio of n (H3PO4) to n (NH3) is 1 to 2, then H3PO4 + 2NH3 = (NH4) 2HPO4 is ammonium hydrogen phosphate;
9) Find the amount of substance (NH4) 2HPO4: n ((NH4) 2HPO4) = n (H3PO4) = 1 mol;
10) Find the mass of the substance (NH4) 2HPO4: m ((NH4) 2HPO4) = n ((NH4) 2HPO4) * Mr ((NH4) 2HPO4) = 1 * 132 = 132 g;
11) Find the mass of the resulting solution: m solution = m (P2O5) + m (H2O) + m (NH3) = 71 + 495 + 34 = 600 g;
12) Find the mass fraction of (NH4) 2HPO4 in the solution: ω ((NH4) 2HPO4) = m ((NH4) 2HPO4) * 100% / m solution = 132 * 100% / 600 = 22%.
Answer: The mass fraction of ammonium hydrogen phosphate in the solution is 22%.