Đáp án:

`↓` 

Giải thích các bước giải:

Ta có `PTHH:`

`Zn+2HCl->ZnCl_2+H_2`

`a)`

`b)`

`n_{H_2}=n_{Zn}=13/65=0,2(mol)`

`=>V_{H_2}=0,2xx22,4=4,48(lít)`

`c)`

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

`=>m_{ct_{HCl}}=0,4xx36,5=14,6(gam)`

`=>m_{dd_{HCl}}=(14,6xx100)/(14,6)=100(gam)`

`=>m_{dd_{spứ}}=100+13-0,2xx2=112,6(gam)`

`=>C%_{ZnCl_2}=(136xx0,2xx100)/(112,6)~~24,6(%)`

`#dkt`

Em tham khảo nha:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{{13}}{{65}} = 0,2\,mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
{n_{HCl}} = 2{n_{Zn}} = 0,4\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,4 \times 36,5}}{{14,6\% }} = 100g\\
{m_{{\rm{dd}}spu}} = 13 + 100 - 0,2 \times 2 = 112,6g\\
{C_\% }ZnC{l_2} = \dfrac{{0,2 \times 136}}{{112,6}} \times 100\%  = 24,16\% 
\end{array}\)