1)

a) \(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)

b) \(2K + 2{H_2}O\xrightarrow{{}}2KOH + {H_2}\)

c) \(Ca + 2{H_2}O\xrightarrow{{}}Ca{(OH)_2} + {H_2}\)

d) \(F{e_2}{O_3} + 3CO\xrightarrow{{{t^o}}}2Fe + 3C{O_2}\)

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

g) \(F{e_3}{O_4} + 4CO\xrightarrow{{{t^o}}}3Fe + 4C{O_2}\)

h) \(F{e_3}{O_4} + 8HCl\xrightarrow{{}}FeC{l_2} + 2FeC{l_3} + 4{H_2}O\)

i) \(F{e_2}{O_3} + 2Al\xrightarrow{{{t^o}}}2Fe + A{l_2}{O_3}\)

k) \(3F{e_3}{O_4} + 8Al\xrightarrow{{{t^o}}}9Fe + 4A{l_2}{O_3}\)

2)

Phản ứng xảy ra:

\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)

Ta có: 

\({n_{Na}} = \frac{{4,6}}{{23}} = 0,2{\text{ mol = }}{{\text{n}}_{NaOH}} \to {m_{NaOH}} = 0,2.40 = 8{\text{ gam;}}{{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,1{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

Khử CuO bằng lượng hidro trên

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

Ta có: \({n_{CuO}} = \frac{{24}}{{64 + 16}} = 0,3{\text{ mol > }}{{\text{n}}_{{H_2}}}\) nên hiệu suất tính theo hidro.

\({n_{{H_2}{\text{ phản ứng}}}} = 0,1.90\%  = 0,09{\text{ mol = }}{{\text{n}}_{Cu}} \to {m_{Cu}} = 0,09.64 = 5,76{\text{ gam}}\)

3)

Phản ứng xảy ra:

\(Ca + 2{H_2}O\xrightarrow{{}}Ca{(OH)_2} + {H_2}\)

Ta có: 

\({n_{Ca}} = \frac{4}{{40}} = 0,1{\text{ mol = }}{{\text{n}}_{{H_2}}} = {n_{Ca{{(OH)}_2}}}\)

\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít;}}{{\text{m}}_{Ca{{(OH)}_2}}} = 0,1.(40 + 17.2) = 7,4{\text{ gam}}\)

Điều chế lượng hidro trên

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

\( \to {n_{{H_2}O}} = \frac{{{n_{{H_2}}}}}{{80\% }} = \frac{{0,1}}{{80\% }} = 0,125{\text{ mol}} \to {{\text{m}}_{{H_2}O}} = 0,125.18 = 2,25{\text{ gam}}\)

Đáp án:

Giải thích các bước giải:

 a) Na+H2O→Na OH+H2

b) 2K+2H2O→2KOH+H2

c) Ca+2H2O→Ca(OH)2+H2

d) Fe2O3+3CO→2Fe+3CO2

e)CuO+H2→Cu+H2O

g)Fe3O4+ 4CO → 3Fe + 4CO₂

_{h) Fe3O4 + 8HCl → 2FeCl3 + FeCl2 + 4H2O}

_{i) 2Al+Fe2O3⟶Al2O3+2Fe}

_{k)8Al+3Fe3O4⟶4Al2O3+9Fe}

bai 2

nNa=4,6/23=0,2(mol)

a)PTHH:

2Na+2H2O->H2+2NaOH

2       2              1          2 

0,2                

b)nH2=0,1mol

VH2=0,1.22,4=2,24(lít)

c)mNaOH=0,2.(23+16+1)=0,2.40=8(g)

d) CuO+H2→Cu+H2O

nCu=0,1 mol

mCu=0,1.64=6,4g

bai 3

2Ca+2H2O→2Ca(OH)2+H2

1       2          1              1
0,1

nCa=4/40=0,1mol

nH2=0,1mol

VH2=0,1.22,4=2,24l

nCa(OH)2=0,1mol

mCaOH=0,1.57=5,7g

2H2+O2→2H2O

nH2O=0,1mol

mH2O=0,1.18=1,8g