J.R. S. answered 04/29/20
Ph.D. University Professor with 10+ years Tutoring Experience
I'll assume the Zn is reacting with HCl as follows:
Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g) ... balanced equation
20.8 g ZnCl2 x 1 mol ZnCl2/136 g x1 mol Zn/mol ZnCl2 x 65.4 g Zn/mol = 10.0 g Zn needed (3 sig. figs.)