Đáp án:

$a\big)Zn+2HCl\to ZnCl_2+H_2$ 

$b\big)C\%_{HCl}=3,65\%$

$c\big)V_{H_2(đktc)}=2,24(l)$

$d\big)C\%_{ZnCl_2}\approx 6,59\%$

Giải thích các bước giải:

$a\big)Zn+2HCl\to ZnCl_2+H_2$

$b\big)$

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

Theo PT: $n_{HCl}=2n_{Zn}=0,2(mol)$

$\to C\%_{HCl}=\frac{0,2.36,5}{200}.100\%=3,65\%$

$b\big)$

Theo PT: $n_{H_2}=n_{Zn}=0,1(mol)$

$\to V_{H_2(đktc)}=0,1.22,4=2,24(l)$

$c\big)$

Theo PT: $n_{ZnCl_2}=n_{Zn}=0,1(mol)$

Ta có $m_{dd\,spu}=6,5+200-0,1.2=206,3(g)$

$\to C\%_{ZnCl_2}=\frac{0,1.136}{206,3}.100\%\approx 6,59\%$

Em tham khảo nha:

\(\begin{array}{l}
d)\\
{m_{{\rm{dd}}spu}} = {m_{Zn}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 6,5 + 200 - 0,1 \times 2 = 206,3g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,1\,mol\\
{C_\% }ZnC{l_2} = \dfrac{{0,1 \times 136}}{{206,3}} \times 100\%  = 6,59\% 
\end{array}\)