Calcium phosphide weighing 72.8 g reacts with water to form calcium hydroxide and hydrogen phosphide

Calcium phosphide weighing 72.8 g reacts with water to form calcium hydroxide and hydrogen phosphide (phosphine). The resulting gas is burned, and the resulting phosphorus (V) oxide is dissolved in 100 ml of a 25% sodium hydroxide solution having a density of 1.28 g / ml until the oxide is completely dissolved. Determine what salt is formed and what is its mass concentration in the resulting solution.

Given:
m (Ca3P2) = 72.8 g
V (NaOH) = 100 ml
ω (NaOH) = 25%
ρ solution (NaOH) = 1.28 g / ml

To find:
ω (salt) -?

Solution:
1) Ca3P2 + 6H2O => 3Ca (OH) 2 + 2PH3 ↑;
2PH3 + 4O2 => P2O5 + 3H2O;
2) n (Ca3P2) = m / M = 72.8 / 182 = 0.4 mol;
3) n (PH3) = 2 * n (Ca3P2) = 2 * 0.4 = 0.8 mol;
4) n (P2O5) = n (PH3) / 2 = 0.8 / 2 = 0.4 mol;
5) m solution (NaOH) = ρ * V = 1.28 * 100 = 128 g;
6) m (NaOH) = ω * m solution / 100% = 25% * 128/100% = 32 g;
7) n (NaOH) = m / M = 32/40 = 0.8 mol;
8) P2O5 + 2NaOH + H2O => 2NaH2PO4;
9) n (NaH2PO4) = n (NaOH) = 0.8 mol;
10) m (NaH2PO4) = n * M = 0.8 * 120 = 96 g;
11) m (P2O5) = n * M = 0.4 * 128 = 56.8 g;
12) m solution = m (P2O5) + m (NaOH) = 56.8 + 128 = 184.8 g;
13) ω (NaH2PO4) = m * 100% / m solution = 96 * 100% / 184.8 = 51.95%.

Answer: The mass fraction of NaH2PO4 is 51.95%.