@$\text{Ly}$

$\text{Câu 1:}$

$a)$ $n_{Zn}$ $=$ $\dfrac{19,5}{65}$ $=$ $0,3$ $mol$

PTHH: $Zn$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2↑$

Theo PT, $n_{ZnCl_2}$ $=$ $n_{Zn}$ $=$ $0,3$ $mol$

$m_{ZnCl_2}$ $=$ $0,3.136=40,8$ $(g)$

$b)$ Tương tự, ta có: $n_{HCl}$ $=$ $2.n_{Zn}$ $=$ $0,6$ $mol$

$m_{HCl}$ $=$ $0,6.36,5=21,9$ $(g)$

`C%_(HCl` $=$ $\dfrac{21,9}{200}$`.100%` $=$ $10,95$`%`

$\text{Câu 2:}$

$a)$ $n_{Al}$ $=$ $\dfrac{8,1}{27}$ $=$ $0,3$ $mol$

PTHH: $2Al$ $+$ $6HCl$ `->` $2AlCl_3$ $+$ $3H_2↑$

Theo PT, $n_{HCl}$ $=$ $3.n_{Al}$ $=$ $0,9$ $mol$

$V_{HCl}$ $=$ $\dfrac{0,9}{1,5}$ $=$ $0,6$ $(l)$

$b)$ Tương tự, $n_{AlCl_3}$ $=$ $n_{Al}$ $=$ $0,3$ $mol$

$m_{AlCl_3}$ $=$ $0,3.133,5=40,05$ $(g)$

 

1)

Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

Ta có:

\({n_{Zn}} = \frac{{19,5}}{{65}} = 0,3{\text{ mol = }}{{\text{n}}_{ZnC{l_2}}}\)

\( \to {m_{ZnC{l_2}}} = 0,3.(65 + 35,5.2) = 40,8{\text{ gam}}\)

Ta có:

\({n_{HCl}} = 2{n_{Zn}} = 0,3.2 = 0,6{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,6.36,5 = 21,9{\text{ gam}}\)

\( \to C{\% _{HCl}} = \frac{{21,9}}{{200}}.100\%  = 10,95\% \)

2)

Phản ứng xảy ra:

\(2{\text{A}}l + 6HCl\xrightarrow{{}}2{\text{A}}lC{l_3} + 3{H_2}\)

Ta có:

\({n_{Al}} = \frac{{8,1}}{{27}} = 0,3{\text{ mol}} \to {{\text{n}}_{HCl}} = 3{n_{Al}} = 0,3.3 = 0,9{\text{ mol}}\)

\( \to V={V_{dd\;{\text{HCl}}}} = \frac{{0,9}}{{1,5}} = 0,6{\text{ lít = 600 ml}}\)

Ta có:

\({n_{AlC{l_3}}} = {n_{Al}} = 0,3{\text{ mol}}\)

\( \to {m_{AlC{l_3}}} = 0,3.(27 + 35,5.3) = 40,05{\text{ gam}}\)