Đáp án + Giải thích các bước giải:

$a\big)$

$n_{Zn}=\frac{65}{65}=1(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

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

$\to V_{H_2(đktc)}=1.22,4=22,4(l)$

$b\big)$

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

$\to m_{ZnCl_2}=1.136=136(g)$

$c\big)$

$m_{dd\,spu}=65+93,7-1.2=156,7(g)$

$\to C\%_{ZnCl_2}=\frac{136}{156,7}.100\%\approx 86,79\%$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1\,mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
b)\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,1\,mol\\
{m_{ZnC{l_2}}} = 0,1 \times 136 = 13,6g\\
c)\\
{m_{{\rm{dd}}spu}} = 6,5 + 93,7 - 0,1 \times 2 = 100g\\
{C_\% }ZnC{l_2} = \dfrac{{13,6}}{{100}} \times 100\%  = 13,6\% 
\end{array}\)