Question

Answer 1

nZn = 8.125/65 = 0.125 (mol) 

nHCl = 18.25/36.5 = 0.5 (mol) 

Zn + 2HCl => ZnCl2 + H2 

0.125...0.25......0.125....0.125

=> HCl dư 

VH2 = 0.125*22.4 = 2.8 (l) 

mZnCl2 = 0.125*136 = 17 (g) 

 

Answer 2

a) nZn= 0,125(mol)

nHCl=0,5(mol)

PTHH: Zn + 2 HCl -> ZnCl2 + H2

Ta có: 0,125/1 < 0,5/2

=> HCl dư, Zn hết, tính theo nZn

=>nH2=nZnCl2=nZn=0,125(mol)

=> V(H2,đktc)=0,125.22,4=2,8(l)

b) mZnCl2=136.0,125=17(g)