Đáp án:

\(\begin{array}{l}
A){m_{{\rm{dd}}HCl}} = 48,67g\\
B)\\
{m_{{H_2}}} = 0,4g\\
{V_{{H_2}}} = 4,48l\\
C)C{\% _{ZnC{l_2}}} = 44,4\% \\
D)C{M_{ZnC{l_2}}} = 4M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = 0,2mol\\
A)\\
{n_{HCl}} = 2{n_{Zn}} = 0,4mol\\
 \to {m_{HCl}} = 14,6g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{14,6}}{{30\% }} \times 100\%  = 48,67g
\end{array}\)

\(\begin{array}{l}
B)\\
{n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
 \to {m_{{H_2}}} = 0,4g\\
 \to {V_{{H_2}}} = 4,48l
\end{array}\)

\(\begin{array}{l}
C)\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,2mol\\
 \to {m_{ZnC{l_2}}} = 27,2g\\
{m_{{\rm{dd}}}} = {m_{Zn}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 61,27g\\
 \to C{\% _{ZnC{l_2}}} = \dfrac{{27,2}}{{61,27}} \times 100\%  = 44,4\% 
\end{array}\)

\(\begin{array}{l}
D)\\
{V_{ZnC{l_2}}} = \dfrac{{61,27}}{{1,163}} = 52,68ml = 0,05l\\
C{M_{ZnC{l_2}}} = \dfrac{{0,2}}{{0,05}} = 4M
\end{array}\)

`n_{Zn}=\frac{13}{65}=0,2(mol)`

`a)` `Zn+2HCl\to ZnCl_2+H_2`

`n_{HCl}=2n_{Zn}=0,4(mol)`

`\to m_{dd\ HCl}=\frac{100%.0,4.36,5}{30%}\approx 48,67g`

`b)` `n_{H_2}=n_{Zn}=0,2(mol)`

`\to m_{H_2}=2.0,2=0,4g`

` V_{H_2}=0,2.22,4=4,48(l)`

`c)` `m_{\text{dd spu}}=m_{Zn}+m_{\text{dd HCl}}-m_{H_2}`

`m_{\text{dd spu}}=13+48,67-0,2.2=61,27g`

`C%_{ZnCl_2}=\frac{136.100%.0,2}{61,27}\approx 44,44%`

`d)` `V_{HCl}=\frac{m_{dd}}{D}=\frac{48,67}{1,163}=41,84ml=0,04184(l)`

`=> C_{M\ ZnCl_2}=\frac{0,2}{0,04184}=4,8M`