Đáp án:

 a) VH2=13,44 lít

b) mFe=18,816 gam

Giải thích các bước giải:

 a)Phản ứng xảy ra :

\(2{H_2}O\xrightarrow{{}}2{H_2} + {O_2}\)

Ta có:  \({n_{{H_2}O}} = \frac{{10,8}}{{18}} = 0,6{\text{ mol = }}{{\text{n}}_{{H_2}}} \to {V_{{H_2}}} = 0,6.22,4 = 13,44{\text{ lít}}\)

b) \(F{e_2}{O_3} + 3{H_2}\xrightarrow{{}}2Fe + 3{H_2}O\)

Ta có:  \({m_{F{e_2}{O_3}}} = 38,4.70\%  = 26,88{\text{ gam}} \to {{\text{n}}_{F{e_2}{O_3}}} = \frac{{26,88}}{{56.2 + 16.3}} = 0,168{\text{ mol < }}\frac{1}{3}{n_{{H_2}}}\)

Do đó H2 dư.

\(\to {n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,168.2 = 0,336{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,336.56 = 18,816{\text{ gam}}\)

$pthh:$

$2H2O → 2H2 + O2$

 $nH2O=10,8/18=0,6mol$

$a/$

$theo$ $pt :$

$nH2=nH2O=0,6mol$

$⇒V_{H_{2}}=0,6.22,4=13,44l$

$b/$

$pthh :$

$Fe2O3 + 3H2 → 2Fe+3H2O$

$mFe3O4=70\%.38,4=26,88g$

$⇒nFe3O4=26,88/232=0,168mol$

ta thấy 

$0,168mol < \frac{1}{3}n_{H_{2}}$

$⇒H_{2} dư$

$⇒nFe=2nFe2O3=2.0,168=0,336mol$

$⇒mFe=0,336.56=18,816g$