Đáp án:
\(\begin{array}{l}
{m_{Cu{{(OH)}_2}}} = 9,8g\\
C{M_{NaOH(dư)}} = C{M_{N{a_2}S{O_4}}} = 0,25M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2NaOH + C{\rm{uS}}{O_4} \to N{a_2}S{O_4} + Cu{(OH)_2}\\
{n_{{\rm{CuS}}{O_4}}} = 0,1mol\\
{n_{NaOH}} = 0,3mol\\
\to {n_{NaOH}} > {n_{{\rm{CuS}}{O_4}}} \to {n_{NaOH}}dư\\
{n_{Cu{{(OH)}_2}}} = {n_{{\rm{CuS}}{O_4}}} = 0,1mol\\
\to {m_{Cu{{(OH)}_2}}} = 9,8g\\
{n_{N{a_2}S{O_4}}} = {n_{{\rm{CuS}}{O_4}}} = 0,1mol\\
{n_{NaOH}} = 2{n_{{\rm{CuS}}{O_4}}} = 0,2mol\\
\to {n_{NaOH(dư)}} = 0,1mol\\
\to C{M_{NaOH(dư)}} = C{M_{N{a_2}S{O_4}}} = \dfrac{{0,1}}{{0,15 + 0,25}} = 0,25M\\
\end{array}\)