Đáp án:
\(\begin{array}{l}
b)\\
{m_{ddNaOH}} = 120g\\
c)\\
C{\% _{N{a_2}S{O_4}}} = 9,45\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuS{O_4} + 2NaOH \to N{a_2}S{O_4} + Cu{(OH)_2}\\
{m_{CuS{O_4}}} = \dfrac{{120 \times 20}}{{100}} = 24g\\
{n_{CuS{O_4}}} = \dfrac{{24}}{{160}} = 0,15mol\\
{n_{NaOH}} = 2{n_{CuS{O_4}}} = 0,3mol\\
{m_{NaOH}} = 0,3 \times 40 = 12g\\
{m_{ddNaOH}} = \dfrac{{12 \times 100}}{{10}} = 120g\\
c)\\
{n_{Cu{{(OH)}_2}}} = {n_{CuS{O_4}}} = 0,15mol\\
{m_{Cu{{(OH)}_2}}} = 0,15 \times 98 = 14,7g\\
{m_{ddspu}} = 120 + 120 - 14,7 = 225,3g\\
{n_{N{a_2}S{O_4}}} = {n_{CuS{O_4}}} = 0,15mol\\
{m_{N{a_2}S{O_4}}} = 0,15 \times 142 = 21,3g\\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{21,3}}{{225,3}} \times 100\% = 9,45\%
\end{array}\)
