$\text{2) Sơ đồ chuyển hoá:}$

$H_2O\xrightarrow{(1)}O_2\xrightarrow{(2)}NaOH \xrightarrow{(3)}Na_2O\xrightarrow{(4)}NaCl$

$\text{PTHH:}$

$(1) 2H_2O\xrightarrow{đp}2H_2+O_2$

$(2) 2Na+O_2\xrightarrow{t^0}2Na_2O$

$(3) 2NaOH\xrightarrow{t^0} Na_2O+H_2O$

 $(4) Na_2O+2HCl\to 2NaCl+H_2O$

$\text{3)}

$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

$\text{PTHH:}$

$2Na+2H_2O\to 2NaOH+H_2$

$\text{Ta có:} n_{Na}=2n_{H_2}=0,6(mol)$

$\to m=m_{Na}=0,6.23=13,8(g)$

2)

Sơ đồ:

\({H_2}O\xrightarrow{{}}{O_2}\xrightarrow{{}}N{a_2}O\xrightarrow{{}}NaOH\xrightarrow{{}}NaCl\)

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

\(4Na + {O_2}\xrightarrow{{{t^o}}}2N{a_2}O\)

\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)

\(NaOH + HCl\xrightarrow{{}}NaCl + {H_2}O\)

3)

Phản ứng xảy ra:

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

Ta có:

\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\)

Theo phản ứng:

\({n_{Na}} = 2{n_{{H_2}}} = 0,3.2 = 0,6{\text{ mol}}\)

\( \to m = {m_{Na}} = 0,6.23 = 13,8{\text{ gam}}\)