@$\text{Ly}$

$a)$ $n_{Fe}$ $=$ $\dfrac{5,6}{56}$ $=$ $0,1$ $mol$

PTHH: $Fe$ $+$ $2HCl$ `->` $FeCl_2$ $+$ $H_2↑$

Theo PT, $n_{FeCl_2}$ $=$ $n_{Fe}$ $=$ $0,1$ $mol$

$m_{FeCl_2}$ $=$ $0,1.127=12,7$ $(g)$

$b)$ $FeCl_2$: Sắt $(II)$ clorua

$⇒$ PƯ trên là PƯ thế

$c)$ Ta có: $n_{H_2}$ $=$ $n_{Fe}$ $=$ $0,1$ $mol$

$V_{H_2}$ $=$ $0,1.22,4=2,24$ $(l)$

$d)$ Tương tự, $n_{HCl}$ $=$ $2.n_{Fe}$ $=$ $0,2$ $mol$

$m_{HCl}$ $=$ $0,2.36,5=7,3$ $(g)$

`C%_(HCl` $=$ $\dfrac{7,3}{14,6}$`.100%` $=$ $50$`%`

$d)$ PTHH: $CuO$ $+$ $H_2$ $\xrightarrow[]{t^o}$ $Cu$ $+$ $H_2O$

Theo PT trên, $n_{H_2O}$ $=$ $n_{H_2}$ $=$ $0,1$ $mol$

$m_{H_2O}$ $=$ $0,1.18=1,8$ $(g)$

Đáp án:

\({m_{F{\text{e}}C{l_2}}} = 12,7{\text{ gam}}\)

\({n_{{H_2}}} = 2,24{\text{ lít}}\)

\(C{\% _{HCl}}  = 50\% \)

\({{\text{m}}_{{H_2}O}} = 1,8{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(F{\text{e}} + 2HCl\xrightarrow{{}}F{\text{e}}C{l_2} + {H_2}\)

Ta có:

\( \to {n_{F{\text{e}}}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol = }}{{\text{n}}_{F{\text{e}}C{l_2}}} = {n_{{H_2}}}\)

\( \to {m_{F{\text{e}}C{l_2}}} = 0,1.(56 + 35,5.2) = 12,7{\text{ gam}}\)

Muối tạo ra là \(FeCl_2\) (sắt (II) clorua). Phản ứng trên là phản ứng thế.

\( \to {n_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

\({n_{HCl}} = 2{n_{F{\text{e}}}} = 0,1.2 = 0,2{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,2.36,5 = 7,3{\text{ gam}}\)

\( \to C{\% _{HCl}} = \frac{{7,3}}{{14,6}}.100\%  = 50\% \)

Cho lượng \(H_2\) trên khử \(CuO\)

\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)

Ta có:

\({n_{{H_2}O}} = {n_{{H_{ 2}}}} = 0,1{\text{ mol}} \to {{\text{m}}_{{H_2}O}} = 0,1.18 = 1,8{\text{ gam}}\)