Em tham khảo nha:
\(\begin{array}{l}
2)\\
{n_{NaOH}} = 1,5 \times 1,5 = 2,25\,mol\\
{V_{{\rm{dd}}NaOH1\,M}} = \dfrac{{2,25}}{1} = 2,25l\\
{V_{{H_2}O}} = 2,25 - 1,5 = 0,75l\\
3)\\
a)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
b)\\
{n_{Na}} = \dfrac{{13,8}}{{23}} = 0,6\,mol\\
{n_{{H_2}}} = \dfrac{{0,6}}{2} = 0,3\,mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
c)\\
{m_{{\rm{dd}}spu}} = 13,8 + 50 - 0,3 \times 2 = 63,2g\\
{n_{NaOH}} = {n_{Na}} = 0,6\,mol\\
{C_\% }NaOH = \dfrac{{0,6 \times 40}}{{63,2}} \times 100\% = 37,97\% \\
4)\\
a)\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
b)\\
{n_{{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
hh:Fe(a\,mol),S(b\,mol)\\
\left\{ \begin{array}{l}
56a + 32b = 20\\
\frac{{2a}}{3} + b = 0,3
\end{array} \right.\\
\Rightarrow a = 0,3;b = 0,1\\
\% {m_{Fe}} = \dfrac{{0,3 \times 56}}{{20}} \times 100\% = 84\% \\
\% {m_S} = 100 - 84 = 16\%
\end{array}\)
