Đáp án:

\({{\text{V}}_{{H_2}}} = 5,04{\text{ lít}}\)

\({{\text{m}}_{Fe}} = 8,4{\text{ gam}}\)

Giải thích các bước giải:

Phản ứng xảy ra:

\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)

Ta có: 

\({n_{F{e_2}{O_3}}} = \frac{{12}}{{56.2 + 16.3}} = \frac{{12}}{{56.2 + 16.3}} = 0,075{\text{ mol}}\)

Ta có: 

\({n_{{H_2}}} = 3{n_{F{e_2}{O_3}}} = 0,075.3 = 0,225{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,225.22,4 = 5,04{\text{ lít}}\)

Ta có: 

\({n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,15{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,15.56 = 8,4{\text{ gam}}\)

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

Ta có: 

\({n_{Zn}} = \frac{{22,57}}{{65}} = 0,347{\text{ mol;}}{{\text{n}}_{HCl}} = \frac{{29,2}}{{36,5}} = 0,8{\text{ mol > 2}}{{\text{n}}_{Zn}}\)

Vậy HCl dư

\( \to {n_{HCl{\text{ dư}}}} = 0,8 - 0,347.2 = 0,106{\text{ mol}} \to {{\text{m}}_{HCl{\text{ dư}}}} = 0,106.36,5 = 3,869{\text{ gam}}\)

$n_{Fe_2O_3}=12/160=0,075mol$

$PTHH :$

$Fe_2O_3+3H_2\overset{t^o}\to 2Fe+3H_2O$

a/Theo pt :

$n_{H_2}=3.n_{Fe_2O_3}=3.0,075=0,225mol

$⇒V_{H_2}=0,225.22,4=5,04l

$\text{b/Theo pt :}$

$n_{Fe}=2.n_{Fe_2O_3}=2.0,075=0,15mol$

$⇒m_{Fe}=0,15.56=8,4g$

$c/n_{Zn}=22,57/65=0,347mol$

$n_{HCl}=29,2/36,5=0,8mol$

                        $Zn  +  2HCl\to ZnCl_2+H_2↑$

$\text{Theo pt :     1 mol        2 mol}$

$\text{Theo đbài : 0,374mol   0,8mol}$

$\text{⇒Sau pư HCl dư 0,106mol}$

$⇒m_{HCl dư}=0,106.36,5=3,896g$