\(m_{CuSO_4}=\dfrac{mdd\cdot C\%}{100\%}=\dfrac{24\cdot40\%}{100\%}=9,6\left(g\right)\)
\(n_{CuSO_4}=\dfrac{9,6}{160}=0,06\left(mol\right)\)
PTHH:
\(CuSO_4+2NaOH->Cu\left(OH\right)_2+Na_2SO_4\)
0,06 --> 0,12 0,06 0,06 (mol)
\(m_{NaOH}=0,12\cdot40=4,8\left(g\right)\)
\(m_{ddNaOH}=\dfrac{m_{NaOH}\cdot100\%}{C\%}=\dfrac{4,8\cdot100\%}{24\%}=20\left(g\right)\)
\(m_{Cu\left(OH\right)_2}=0,06\cdot98=5,88\left(g\right)\)
\(m_{Na_2SO_4}=0,06\cdot142=8,52\left(g\right)\)
\(m_{ddsau}=m_{ddNaOH}+m_{ddCuSO_4}-m_{Cu\left(OH\right)_2}\)
\(m_{ddsau}=20+24-5,88=38,12\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{8,52}{38,12}\cdot100\%\approx22,35\%\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
0,06 --> 0,06 (mol)
\(m_{CuO}=0,06\cdot80=4,8\left(g\right)\)