Calculate the mass of the acid obtained by the interaction of 37.5 g of a 40% solution of methanal with copper hydroxide (ll).

Given:
m solution (HCHO) = 37.5 g
ω (HCHO) = 40%

To find:
m (acid) -?

1) HCHO + Cu (OH) 2 => HCOOH + Cu ↓ + H2O;
2) m (HCHO) = ω (HCHO) * m solution (HCHO) / 100% = 40% * 37.5 / 100% = 15 g;
3) M (HCHO) = Mr (HCHO) = Ar (H) * N (H) + Ar (C) * N (C) + Ar (O) * N (O) = 1 * 2 + 12 * 1 + 16 * 1 = 30 g / mol;
4) n (HCHO) = m (HCHO) / M (HCHO) = 15/30 = 0.5 mol;
5) n (HCOOH) = n (HCHO) = 0.5 mol;
6) M (HCOOH) = Mr (HCOOH) = Ar (H) * N (H) + Ar (C) * N (C) + Ar (O) * N (O) = 1 * 2 + 12 * 1 + 16 * 2 = 46 g / mol;
7) m (HCOOH) = n (HCOOH) * M (HCOOH) = 0.5 * 46 = 23 g.

Answer: The mass of HCOOH is 23 g.