Find the mass of the salt that is formed by the interaction of 15% phosphoric acid weighing 215 grams with magnesium

Given:
ω (H3PO4) = 15%
m solution (H3PO4) = 215 g

To find:
m (salt) -?

1) 2H3PO4 + 3Mg => Mg3 (PO4) 2 + 3H2;
2) m (H3PO4) = ω * m solution / 100% = 15% * 215/100% = 32.25 g;
3) n (H3PO4) = m / M = 32.25 / 98 = 0.33 mol;
4) n (Mg3 (PO4) 2) = n (H3PO4) / 2 = 0.33 / 2 = 0.17 mol;
5) m (Mg3 (PO4) 2) = n * M = 0.17 * 262 = 45 g.

Answer: The mass of Mg3 (PO4) 2 is 45 g.