# In 250 g of a solution with a mass fraction of orthophosphoric acid of 9.8%, 14.2 g of phosphorus (5)

**In 250 g of a solution with a mass fraction of orthophosphoric acid of 9.8%, 14.2 g of phosphorus (5) oxide was dissolved. Determine the mass fraction (%) of phosphoric acid in the resulting solution.**

Given:

m1 solution (Н3РО4) = 250 g

ω1 (Н3РО4) = 9.8%

m2 (P2O5) = 14.2 g

Find:

ω3 (Н3РО4) -?

Solution:

1) P2O5 + 3H2O => 2 H3PO4;

2) M (P2O5) = Mr (P2O5) = Ar (P) * 2 + Ar (O) * 5 = 31 * 2 + 16 * 5 = 142 g / mol;

M (H3PO4) = Mr (H3PO4) = Ar (H) * 3 + Ar (P) + Ar (O) * 4 = 1 * 3 + 31 + 16 * 4 = 98 g / mol;

3) m1 (Н3РО4) = ω1 (Н3РО4) * m1 solution (Н3РО4) / 100% = 9.8% * 250/100% = 24.5 g;

4) n2 (P2O5) = m2 (P2O5) / M (P2O5) = 14.2 / 142 = 0.1 mol;

5) n2 (H3PO4) = n2 (P2O5) * 2 = 0.1 * 2 = 0.2 mol;

6) m2 (H3PO4) = n2 (H3PO4) * M (H3PO4) = 0.2 * 98 = 19.6 g;

7) m3 (H3PO4) = m1 (H3PO4) + m2 (H3PO4) = 24.5 + 19.6 = 44.1 g;

8) m3 solution (Н3РО4) = m1 solution (Н3РО4) + m2 (Р2О5) = 250 + 14.2 = 264.2 g;

9) ω3 (Н3РО4) = m3 (Н3РО4) * 100% / m3 solution (Н3РО4) = 44.1 * 100% / 264.2 = 16.7%.

Answer: The mass fraction of Н3РО4 in the resulting solution is 16.7%.