Giải thích các bước giải:

5,

\(\begin{array}{l}
Na + {H_2}O \to NaOH + \frac{1}{2}{H_2}\\
NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}O\\
{n_{Na}} = \dfrac{{2,3}}{{23}} = 0,1mol
\end{array}\)

27ml nước suy ra có 27g nước

\( \to {m_{{\rm{ddNaOH}}}} = 2,3 + 27 - 0,1 = 29,2g\)

\({n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,05mol\)

Trong 29,2g dung dịch có 0,05mol \({H_2}S{O_4}\)

Suy ra trong 7,3g dung dịch có \({n_{{H_2}S{O_4}}} = \dfrac{{7,3 \times 0,05}}{{29,2}} = 0,0125mol\)

\(\begin{array}{l}
{m_{{H_2}S{O_4}{\rm{dd}}}} = \dfrac{{0,0125 \times 98 \times 100}}{{20}} = 6,125g\\
 \to {V_{{H_2}S{O_4}}} = \dfrac{m}{D} = 0,19ml
\end{array}\)

6,

\(\begin{array}{l}
Na + {H_2}O \to NaOH + \frac{1}{2}{H_2}\\
{n_{Na}} = 0,1mol\\
{n_{{H_2}O}} = 2,66mol\\
{n_{{H_2}O}} > {n_{Na}}
\end{array}\)

\({H_2}O\) dư 

\({n_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,05mol\)

\(\begin{array}{l}
 \to {n_{NaOH}} = {n_{Na}} = 0,1mol \to {m_{NaOH}} = 4g\\
{m_{{\rm{dd}}}} = 2,3 + 47,8 - 0,05 \times 2 = 50g\\
 \to C\% NaOH = \dfrac{4}{{50}} \times 100\%  = 8\% 
\end{array}\)