Em tham khảo nha:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = \dfrac{{20}}{{65}} = \dfrac{4}{{13}}\,mol\\
{n_{{H_2}}} = {n_{Zn}} = \dfrac{4}{{13}}\,mol\\
{m_{{H_2}}} = \dfrac{4}{{13}} \times 2 = 0,615g\\
b)\\
{m_{{\rm{dd}}spu}} = 20 + 180 - 0,615 = 199,385g\\
c)\\
{n_{ZnC{l_2}}} = {n_{Zn}} = \dfrac{4}{{13}}\,mol\\
{C_\% }ZnC{l_2} = \dfrac{{\dfrac{4}{{13}} \times 136}}{{199,385}} \times 100\%  = 21\% 
\end{array}\)

Đáp án + Giải thích các bước giải:

$a)$

$Zn+2HCl\to ZnCl_2+H_2$

$n_{Zn}=\dfrac{20}{65}=\dfrac{4}{13}(mol)$

$n_{H_2}=n_{Zn}=\dfrac{4}{13}(mol)$

$m_{H_2}=\dfrac{4}{13}×2=\dfrac{8}{13}(g)$

$b)$

$m_{dd\ sau}=20+180-\dfrac{8}{13}=199,385(g)$

$c)$

$n_{ZnCl_2}=n_{Zn}=\dfrac{4}{13}(mol)$

$C\%_{ZnCl_2}=\dfrac{\dfrac{4}{13}×136}{199,385}×100\%=21\%$