Đáp án:
b) 8g
c) 38,28 ml
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
b)\\
{n_{{H_2}S{O_4}}} = 0,02 \times 1 = 0,02\,mol\\
{n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,04\,mol\\
{m_{{\rm{dd}}NaOH}} = \dfrac{{0,04 \times 40}}{{20\% }} = 8g\\
c)\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,04\,mol\\
{m_{{\rm{ddK}}OH}} = \dfrac{{0,04 \times 56}}{{5,6\% }} = 40g\\
{V_{{\rm{dd}}KOH}} = \dfrac{{40}}{{1,045}} = 38,28ml
\end{array}\)