In water weighing 50 g, p2o5 weighing 8.0 g was dissolved. Formed h3po4, find the mass fraction of this acid.

1) Let’s write the reaction equation:

P2O5 + 3H2O = 2H3PO4

2) Find the number of reacted substances:

n (P2O5) = m (P2O5) / Mr (P2O5) = 8/142 = 0.056 mol;

n (H2O) = m (H2O) / Mr (H2O) = 50/18 = 2.78 mol;

Water in excess, calculation is carried out on phosphorus oxide (V);

3) Find the mass of acid in the resulting solution:

n (H3PO4) = 2n (P2O5) = 2 * 0.056 = 0.112 mol;

m (H3PO4) = n (H3PO4) * Mr (H3PO4) = 0.112 * 98 = 10.976 g;

4) Find the mass fraction of acid in the solution:

m (p-pa) = m (H2O) + m (P2O5) = 50 + 8 = 58 g;

ω (H3PO4) = m (H3PO4) / m (p-pa) = 10.976 / 58 = 0.189 or 18.9%;

Answer: 18.9%.