Đáp án:

 b) V H2=4,48  lít

c) mHCl=14,6 gam

Giải thích các bước giải:

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

Ta có: \({n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol}}\)

\( \to {n_{Fe}} = {n_{FeC{l_2}}} = {n_{{H_2}}} = 0,2{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lit}}\)

Tính HCl

Cách 1: \({n_{HCl}} = 2{n_{Fe}} = 0,2.2 = 0,4{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)

Cách 2: \({m_{FeC{l_2}}} = 0,2.(56 + 35,5.2) = 25,4{\text{ gam}}\)

BTKL: \({m_{Fe}} + {m_{HCl}} = {m_{FeC{l_2}}} + {m_{{H_2}}} \to 11,2 + {m_{HCl}} = 25,4 + 0,2.2 \to {m_{HCl}} = 14,6{\text{ gam}}\)

$n_{Fe}=11,2/56=0,2mol$

$a/Fe+2HCl→FeCl_2+H_2↑$

$\text{b/Theo pt :}$

$n_{H_2}=n_{Fe}=0,2mol$

$⇒V_{H_2}=0,2.22,4=4,48l$

$\text{c/Theo pt :}$

$n_{HCl}=2.n_{Fe}=0,4mol$

$⇒m_{HCl}=0,4.36,5=14,6g$