Bài 4:

a)

$n_{Zn}$=$\frac{6,5}{65}$ =0,1(mol)

$m_{HCl}$ =100*14,6%=14,6(g)

$n_{HCl}$ =$\frac{14.6}{36,5}$ =0.4(mol

 PTHH:

   Zn+2HCl→ZnCl2 +H2

Ta thấy:

$\frac{n_{Zn}}{1}$=$\frac{0.1}{1}$ <$\frac{n_{HCl}}{2}$ =$\frac{0.4}{2}$

⇒Zn phản ứng hết,HCl dư

Theo PTHH,ta có:

$n_{H2}$=$n_{ZnCl2}=$n_{Zn}$=0.1(mol)

$n_{HCL}$=2*$n_{Zn}$=0.2(mol)

Vậy $V_{H2}$=0.1*24.79=2.479

b)

$m_{dd}$=100+6.5-0.1*2=106.3(g)

$mHCl_{dư}$ =(0.4-0.2)*36.5=7.3(g)

$m_{ZnCl2}$ =136*0.1=13.6(g)

C% ZnCl2=$\frac{13.6*100}{106.3}$ ≈12.8%

C%HCl=$\frac{7.3*100}{106.3}$≈6.87%