Đáp án:

a)

`Zn + 2HCl -> ZnCl_2 + H_2`

b)

`m_{HCl} = (200 . 20)/(100) = 40` `gam`

`n_{HCl} = (40)/(36,5) = 80/73` `mol`

`n_{ZnCl_2} = 1/2 . n_{HCl} = 40/73` `mol`

`m_{ZnCl_2} = 40/73 . 136 = 74,5` `gam`

c)

`n_{H_2} = 1/2 . n_{HCl} = 40/73` `mol`

`V_{H_2} = 40/73 . 22,4 = 12,27` `l`

d)

`n_(Zn) = 1/2 . n_{HCl} = 40/73` `mol`

`m_(Zn) = 40/73 . 65 = 35,6` `gam`

`m_{dd_(pứ)} = 35,6 + 200 - 40/73 . 2 = 234,5` `gam`

`C%_{ZnCl_2} = (74,5)/(234,5) . 100% = 31,76%`

Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{HCl}} = dfrac{{200 \times 20}}{{100 \times 36,5}} = 1,1mol\\
 \to {n_{Zn}} = \dfrac{1}{2}{n_{HCl}} = 0,55mol\\
 \to {m_{Zn}} = 35,75g\\
{n_{{H_2}}} = {n_{Zn}} = 0,55mol\\
 \to {V_{{H_2}}} = 12,32l\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,55mol\\
 \to {m_{ZnC{l_2}}} = 74,8g\\
{m_{{\rm{dd}}}} = {m_{Zn}} + {m_{HCl}}{\rm{dd - }}{{\rm{m}}_{{H_2}}} = 35,75 + 200 - 0,55 \times 2 = 234,65g\\
 \to C{\% _{ZnC{l_2}}} = \dfrac{{74,8}}{{234,65}} \times 100\%  = 31,88\% 
\end{array}\)