\(m_{CuCl_2}=100\times25\%=25\left(g\right)\)
\(\Rightarrow n_{CuCl_2}=\dfrac{25}{135}=\dfrac{5}{27}\left(mol\right)\)
\(m_{KOH}=140\times20\%=28\left(g\right)\)
\(\Rightarrow n_{KOH}=\dfrac{28}{56}=0,5\left(mol\right)\)
PTHH: CuCl2 + 2KOH → 2KCl + Cu(OH)2↓
Ban đầu: \(\dfrac{5}{27}\) ...........0,5........................................... (mol)
Phản ứng: \(\dfrac{5}{27}\) ...........\(\dfrac{10}{27}\) ......................................... (mol)
Sauphảnứng: 0..............\(\dfrac{7}{54}\) ..→ \(\dfrac{10}{27}\) ...........\(\dfrac{5}{27}\) ......... (mol)
a) \(m_{Cu\left(OH\right)_2}=\dfrac{5}{27}\times98=18,15\left(g\right)\)
b) \(m_{dd}saupư=100+140-18,15=221,85\left(g\right)\)
\(m_{KOH}dư=\dfrac{7}{54}\times56=7,26\left(g\right)\)
\(\Rightarrow C\%_{KOH}dư=\dfrac{7,26}{221,85}\times100\%=3,27\%\)