Is 60 g of a 20% formic acid solution enough to dissolve 4 g of zinc containing 2% impurities that do not dissolve in acid?

Given:
m (HCOOH) = 60 g
ω (HCOOH) = 20%
m tech. (Zn) = 4 g
ω approx. = 2%

To find:
m (HCOOH) – Enough?

Decision:
1) 2HCOOH + Zn => Zn (HCOO) 2 + H2 ↑;
2) m (HCOOH) = ω (HCOOH) * m solution (HCOOH) / 100% = 20% * 60/100% = 12 g;
3) n (HCOOH) = m (HCOOH) / M (HCOOH) = 12/46 = 0.26 mol;
4) ω (Zn) = 100% – ω approx. = 100% – 2% = 98%;
5) m clean. (Zn) = ω (Zn) * m tech. (Zn) / 100% = 98% * 4/100% = 3.92 g;
6) n (Zn) = m pure. (Zn) / M (Zn) = 3.92 / 65 = 0.06 mol;
7) n must. (HCOOH) = n (Zn) * 2 = 0.06 * 2 = 0.12 mol;
8) n (HCOOH)> n should (HCOOH).

Answer: HCOOH is enough.