nZn=32,5/65=0,5(mol)
nHCl=150/35,5=4,1(mol)
Zn+2HCl--->ZnCl2+H2
1___2
0,5___4,1
Ta có: 0,5/1<4,1/2
=>HCl dư
Theo pt: nZnCl2=nZn=0,5(mol)
=>mZnCl2=0,5/136=68(g)
Theo pt: nH2=nZn=0,5(mol)
=>VH2=0,5.22,4=11,2(l)
mdd=32,5+150-0,5.2=181,5(g)
C%ZnCl2=68/181,5.100%~37,5%