n Zn = 15/65 = 3/13(mol)
n S = 6,4/32 = 0,2(mol)
PTHH: Zn + S -to-> ZnS
(mol) 0,2____0,2______0,2__
Tỉ lệ: (3/13)/1 > 0,2/1 -> Zn dư 3/13-0,2= 2/65(mol)
m hh = 2/65.65 + 0,2.97 = 21,4(g)
%m Zn dư =(2/65 . 65 )/21,4.100%=9,3(%)
%m ZnS = 100-9,3=90,7(%)