$$\begin{gathered} n_{CuS{O}_4}=0,5 mol \\ a, \left[C{u}^{2+}\right]=\left[S{O_4}^{2-}\right]=\frac{0,5}{0,5}=1M \\ b, C{u}^{2+}+2O{H}^{-}->Cu{\left(OH\right)}_2 \\ 0,5 1 \left(mol\right) \\ =>V=\frac{1}{0,5}=2 lít \\ Vậy V=2 lít \\ c, B{a}^{2+}+S{O_4}^{2-}->BaS{O}_4 \\ 0,5 0,5 \left(mol\right) \\ =>V=\frac{0,5}{0,25}=2 lít \\ Vậy V=2 lít \end{gathered}$$.

nCuSO4=0,5(mol)nCuSO4=0,5(mol)

$CuS{O}_4(0,5)--->C{u}^{2+}(0,5)+S{O_4}^{2-}(0,5)$

---->$\left\{\begin{array}{l}\left[C{u}^{2+}\right]=\frac{0,5}{0,5}=1M\\ \left[S{{O_4}^{2+}}_{}\right]=\frac{0,5}{0,5}=1M\end{array}\right.$

b)

b

Cu2+(0,5)+2OH−(1)−−−>Cu(OH)2↓Cu2+(0,5)+2OH−(1)−−−>Cu(OH)2↓

nOH−=0,1(mol)nOH−=0,1(mol)

⇒VddKOH=10,5=2(l)⇒VddKOH=10,5=2(l)

c)c)

Ba2+(0,5)+SO2−4(0,5)−−−>BaSO4↓Ba2+(0,5)+SO42−(0,5)−−−>BaSO4↓

nBa2+=0,5(mol)nBa2+=0,5(mol)

⇒VddBaCl2=0,50,25=2(l)