Em tham khảo nha :
\(\begin{array}{l}
12)\\
a)\\
{n_{{H_2}S{O_4}}} = 0,02 \times 1 = 0,02mol\\
{H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,04mol\\
{m_{NaOH}} = 0,04 \times 40 = 1,6g\\
{m_{ddNaOH}} = \dfrac{{1,6 \times 100}}{{20}} = 8g\\
b)\\
{H_2}S{O_4} + 2KOH \to {K_2}S{O_4} + 2{H_2}O\\
{n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,04mol\\
{m_{KOH}} = 0,04 \times 56 = 2,24g\\
{m_{ddKOH}} = \dfrac{{2,24 \times 100}}{{5,6}} = 40g\\
{V_{KOH}} = \dfrac{{40}}{{1,045}} = 38,28ml\\
13)\\
a)\\
CuS{O_4} + 2NaOH \to N{a_2}S{O_4} + Cu{(OH)_2}\\
{m_{CuS{O_4}}} = \dfrac{{64 \times 10}}{{100}} = 6,4g\\
{n_{CuS{O_4}}} = \dfrac{{6,4}}{{160}} = 0,04mol\\
{n_{Cu{{(OH)}_2}}} = {n_{CuS{O_4}}} = 0,04mol\\
{m_{Cu{{(OH)}_2}}} = 0,04 \times 98 = 3,92g\\
b)\\
{n_{NaOH}} = 2{n_{CuS{O_4}}} = 0,08mol\\
{m_{NaOH}} = 0,08 \times 40 = 3,2g\\
{m_{ddNaOH}} = \dfrac{{3,2 \times 100}}{8} = 40g\\
{V_{NaOH}} = \dfrac{{40}}{{1,056}} = 37,88ml\\
c)\\
{n_{N{a_2}S{O_4}}} = {n_{CuS{O_4}}} = 0,04mol\\
{m_{N{a_2}S{O_4}}} = 0,04 \times 142 = 5,68g\\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{5,68}}{{64 + 40 - 3,92}} \times 100\% = 5,68\% \\
d)\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,04mol\\
{m_{CuO}} = 0,04 \times 80 = 3,2g
\end{array}\)
