a,ta có phương trình:Zn+2HCl=>ZnCl2+H2
b,ta có mZn=6.5(g)=>nZn=$\frac{6.5}{65}$=0.1(mol)
=>nH2=0.1(mol)=>vH2=0.1*22.4=2.24(lít)
c,nHCl=0.1*2=0.2(mol)
=>mHCl=0.2*36.5=7.3(g)
=>mddHCl=$\frac{7.3*100}{9.6}$=76.04167(g)
ta có nZnCl2=0.1(mol)=>mZnCl2=0.1*(65+35.5*2)=13.6(g)
mddZnCl2=6.5+76.04167-0.1*2= 82.34167(g)
=>C%=$\frac{13.6}{82.34167}$*100=16.52%