Đáp án:

\(\begin{array}{l}
c)\\
{m_{HCl}} = 7,3g\\
d)\\
{m_{Cl}} = 7,1g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{N_{Zn}}:{N_{HCl}}:{N_{ZnC{l_2}}}:{N_{{H_2}}} = 1:2:1:1\\
c)\\
\\
{m_{Zn}} + {m_{HCl}} = {m_{ZnC{l_2}}} + {m_{{H_2}}}\\
{n_{ZnC{l_2}}} = \dfrac{m}{M} = \dfrac{{13,6}}{{136}} = 0,1mol\\
{n_{{H_2}}} = {n_{ZnC{l_2}}} = 0,1mol\\
{m_{{H_2}}} = n \times M = 0,1 \times 2 = 0,2g\\
{m_{Zn}} + {m_{HCl}} = {m_{ZnC{l_2}}} + {m_{{H_2}}}\\
 \Rightarrow {m_{HCl}} = 13,6 + 0,2 - 6,5 = 7,3g\\
d)\\
\% Cl = \dfrac{{2 \times {M_{Cl}}}}{{{M_{ZnC{l_2}}}}} \times 100\%  = \dfrac{{2 \times 35,5}}{{136}} \times 100\%  = 52,2\% \\
{m_{Cl}} = \dfrac{{13,6 \times 52,2}}{{100}} = 7,1g
\end{array}\)

nZn=6.5/65=0.1(mol)

nZnCl2=13.6/136=0.1(mol)

a. PTHH: Zn+2HCl->ZnCl2+H2

b.

Zn:HCl:ZnCl2:H2=1:2:1:1

c.

mZn+mHCl=mZnCl2+mH2

Tính KL của HCl đã dùng:

Theo PT ta có: nHCl=2nZn=0.1*2=0.2(mol)

=>mHCl=0.2*36.5=7.3(gam)

d.

Ta có: mZnCl2=136(g/mol)

mCl=71x13.6/136=7.1(g)