Đáp án:

 b) 41,216l

c) 1342,2g

d) 17,15%

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{{119,5}}{{65}} \approx 1,84\,mol\\
{n_{{H_2}}} = {n_{Zn}} = 1,84\,mol\\
{V_{{H_2}}} = 1,84 \times 22,4 = 41,216l\\
c)\\
{n_{HCl}} = 1,84 \times 2 = 3,68\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{3,68 \times 36,5}}{{10\% }} = 1343,2g\\
d)\\
{m_{{\rm{dd}}spu}} = 119,5 + 1343,2 - 1,84 \times 2 = 1459,02g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 1,84\,mol\\
{C_\% }ZnC{l_2} = \dfrac{{1,84 \times 136}}{{1459,02}} \times 100\%  = 17,15\% 
\end{array}\)

Bài giải :

Câu 7 :

a.

`-n_{Zn}=\frac{119,5}{65}≈1,838(mol)`

  `Zn+2HCl→ZnCl_2+H_2↑`

1,838 →   3,676       1,838         1,838   (mol)      

b.

`-V_{H_2}(đktc)=1,838.2,4=41,1712(l)`

c.

$-m_{chất..tan..HCl}=3,676.36,5=134,174(g)$

`⇒m_{dd..HCl}=\frac{134,174.100%}{10%}=1341,74(g)`

d.

- Dung dịch thu được sau phản ứng: `ZnCl_2`

$-m_{dd..sau..pứ}=m_{Zn}+m_{dd..HCl}-m_{H_2}$

                    $=119,5+1341,74-1,838.2=1457,564(g)$

`-m_{chất..tan..ZnCl_2}=1,838.136=249,968(g)`

`⇒C%ZnCl_2=\frac{249,968}{1457,564}.100%≈17,15%`