Đáp án:
18. B
19. B
20. B
Giải thích các bước giải:
\(\begin{array}{l}
18.\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 0,05mol\\
{n_{{H_2}S{O_4}}} = 0,04mol\\
\to \dfrac{{{n_{NaOH}}}}{2} < {n_{{H_2}S{O_4}}} \to {n_{{H_2}S{O_4}}}dư\\
\to {n_{N{a_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,025mol\\
\to {m_{N{a_2}S{O_4}}} = 3,55g\\
19.\\
2NaOH + {H_3}P{O_4} \to N{a_2}HP{O_4} + 2{H_2}O\\
{n_{NaOH}} = 0,2mol\\
{n_{{H_3}P{O_4}}} = 0,1mol\\
\to \dfrac{{{n_{NaOH}}}}{2} < {n_{{H_3}P{O_4}}} \to {n_{{H_3}P{O_4}}}dư\\
\to {n_{N{a_2}HP{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,1mol\\
\to {m_{N{a_2}HP{O_4}}} = 14,2g\\
20.\\
{V_X} \times (C{M_{KOH}} + 2C{M_{Ba{{(OH)}_2}}}) = 40 \times (C{M_{HCl}} + 2C{M_{{H_2}S{O_4}}})\\
\to {V_X} \times (0,2 + 0,2) = 40 \times (0,75 + 0,5)\\
\to {V_X} = 125ml = 0,125l
\end{array}\)
